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Work & Energy Past Papers
Solved past paper MCQs for Work & Energy from official UHS, NUMS, SZABMU, DUHS, and KMU examinations. Includes verified distractor autopsies and step-by-step cognitive explanations.
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Official printable A4 booklet containing all 99 verified past paper MCQs with unbroken options, 5-column rapid answer key, and Propolis cognitive error autopsies.
A man pulls a trolley through a distance of 50 m by applying a force of 100 N, which makes an angle of 60° with x-axis. Calculate the work done by the man? (\( \cos 60^{\circ} = 0.5 \))
A
2500 J
B
5340 J
C
6430 J
D
7120 J
View Answer & Propolis Autopsy
Correct Key: Option ADiagnostic Explanation
Concept:
Only the component of force acting in the direction of displacement does work.
Formula:
$$ W = Fd \cos \theta $$
Solution:
Force (\( F \)) = 100 N
Displacement (\( d \)) = 50 m
Angle (\( \theta \)) = 60°
\( W = 100 \times 50 \times \cos(60^{\circ}) \).
\( W = 5000 \times 0.5 = 2500 \text{ J} \).
Why other options are incorrect:
If you ignore the angle completely and just multiply 100 x 50, you get 5000 J (not listed). The other options represent arbitrary math errors.
The amount of work required to stop a moving object is equal to:
A
The velocity of the object
B
The change in kinetic energy of the object
C
The mass of the object times its acceleration
D
The mass of the object times its velocity
View Answer & Propolis Autopsy
Correct Key: Option BDiagnostic Explanation
Concept:
This is the fundamental definition of the Work-Energy Theorem.
Formula:
$$ W_{\text{net}} = \Delta K.E = K.E_f - K.E_i $$
Solution:
To stop an object, its final kinetic energy must become zero.
The work done against the object (negative work) removes its kinetic energy.
Therefore, the magnitude of work required is exactly equal to the object's total initial kinetic energy, which represents its change in kinetic energy.
Why other options are incorrect:
Mass \( \times \) acceleration is Force. Mass \( \times \) velocity is Momentum. Work is an energy measure, not a force or momentum measure.
When dolphin leaves the water it has lots of kinetic energy. At its highest point it's energy is:
A
Kinetic energy
B
Potential energy
C
Elastic potential energy
D
Neither kinetic energy nor potential energy
View Answer & Propolis Autopsy
Correct Key: Option BDiagnostic Explanation
Concept:
By the conservation of mechanical energy in projectile/vertical motion, kinetic energy is traded for gravitational potential energy as an object rises.
Formula:
$$ K.E_{\text{initial}} = P.E_{\text{top}} $$
Solution:
When the dolphin jumps straight up, it decelerates due to gravity until its vertical velocity hits zero at the peak of the jump.
At this exact highest point, all of its initial upward Kinetic Energy has been converted into Gravitational Potential Energy.
Why other options are incorrect:
If it jumps at an angle, it retains some horizontal K.E, but the primary energy conversion defining the "highest point" is the maximization of Potential energy. It is not elastic, nor is it zero.
Which of the following force gives rise to ocean tides?
A
Frictional force
B
Gravitational force
C
Earth's magnetic force
D
Nuclear force
View Answer & Propolis Autopsy
Correct Key: Option BDiagnostic Explanation
Concept:
Ocean tides are caused by the differential gravitational pull of massive celestial bodies on the Earth's oceans.
Formula:
$$ F = \frac{GMm}{r^2} $$
Solution:
The Moon and the Sun exert a gravitational pull on the Earth.
Because the oceans are fluid, they bulge outward toward the Moon (and slightly toward the Sun) due to this gravitational attraction.
This gravitational force is exclusively responsible for the rising and falling of ocean tides.
Why other options are incorrect:
Magnetic forces affect charged particles and compasses, not massive bodies of water. Nuclear forces act only at subatomic scales. Friction opposes motion.
According to work energy principle in linear motion, the work done on the body is equal to:
A
Change in P.E
B
Sum of P.E + K.E
C
Change of K.E
D
Zero
View Answer & Propolis Autopsy
Correct Key: Option CDiagnostic Explanation
Concept:
The standard Work-Energy Theorem states that the net work done by all forces on an object equals the change in its kinetic energy.
Formula:
$$ W_{\text{net}} = \Delta K.E = K.E_f - K.E_i $$
Solution:
If a net force acts on a body in linear motion, it accelerates, changing its velocity.
This change in velocity directly translates to a change in kinetic energy.
Hence, Net Work = Change in K.E.
Why other options are incorrect:
Work done against gravity specifically equals the change in P.E, but the generalized Work-Energy principle for linear motion explicitly defines it via K.E.
13.5 is the largest coefficient, making Option C the greatest kinetic energy.
Why other options are incorrect:
Option D seems intuitively large due to the high speed (4v), but the heavier mass (3M) in Option C combined with a high speed (3v squared = 9) overcomes it.
A 2kg object is released from rest 8m above the surface of Earth. During the fall work done against air resistance is 50J just before it hits the surface its speed is:
A
10 m/sec
B
35 m/sec
C
40 m/sec
D
45 m/sec
View Answer & Propolis Autopsy
Correct Key: Option ADiagnostic Explanation
Concept:
Use the Conservation of Energy with non-conservative forces: The initial potential energy is converted into kinetic energy minus the energy lost to air resistance.
Which of the following expression is constant for a freely falling body?
A
\( mgh + mv^2 \)
B
\( mgh = mv^2 \)
C
\( mgh + \frac{1}{2}mv^2 \)
D
\( mgh = -\frac{1}{2}mv^2 \)
View Answer & Propolis Autopsy
Correct Key: Option CDiagnostic Explanation
Concept:
According to the Law of Conservation of Mechanical Energy, in the absence of air resistance, the total mechanical energy of a freely falling body remains constant at every point in its path.
Their sum, \( mgh + \frac{1}{2}mv^2 \), represents the total mechanical energy, which is a conserved constant.
Why other options are incorrect:
Option A misses the \( 1/2 \) coefficient for kinetic energy. Option B and D misstate the conservation law as an equality between instantaneous PE and KE, which is only true at exactly half the initial height.
A body of mass 4 kg is moving a circle of radius 2m. if the body moves round a complete circle, what is the work done by the body?
A
8 J
B
0
C
16 J
D
5 J
View Answer & Propolis Autopsy
Correct Key: Option BDiagnostic Explanation
Concept:
There are two primary reasons work is zero here: displacement logic and force angle logic.
Formula:
$$ W = \vec{F} \cdot \vec{d} = Fd \cos \theta $$
Solution:
Reason 1: Over a complete circle, the start and end points are identical. Total displacement \( \vec{d} = 0 \). Therefore, Work = 0.
Reason 2: Centripetal force acts towards the center, while instantaneous motion is tangential (angle is 90°). Since \( \cos(90^{\circ}) = 0 \), continuous work done is zero at every instant.
Why other options are incorrect:
Options A, C, and D are calculated by blindly multiplying mass, radius, and other numbers given in the prompt, ignoring the vector nature of work.
An object has 1J of P.E. what is the work done in term of height?
A
10J
B
0J
C
0.1J
D
1J
View Answer & Propolis Autopsy
Correct Key: Option DDiagnostic Explanation
Concept:
Energy is fundamentally defined as the capacity to do work.
Formula:
$$ W = \Delta P.E $$
Solution:
If a body has 1 Joule of Gravitational Potential Energy, it means exactly 1 Joule of work was done to lift it to that height.
Conversely, if released, it has the capacity to do exactly 1 Joule of work as it falls back to the reference level.
Therefore, the work equivalence is directly 1 J.
Why other options are incorrect:
Assuming the mass is 1 kg and gravity is 10 m/s² might trick someone into calculating height (h = 0.1m) and picking 0.1J, but the question asks for work done, which matches the energy perfectly.
A body of mass 'm' is moving with velocity 'v'. after a short interval of time its velocity becomes double. How many time its K.E will increase of decrease?
A
2 time increased
B
2 time decreased
C
4 time increased
D
4 time decreased
View Answer & Propolis Autopsy
Correct Key: Option CDiagnostic Explanation
Concept:
Kinetic energy has a quadratic relationship with velocity.
Because 4 is greater than 1, it is a 4 times increase.
Why other options are incorrect:
Option A mistakes the quadratic relationship for a linear one (momentum doubles, but K.E quadruples). Decreasing options are illogical since speed increased.
Option A forgets the \( 1/2 \) factor in the kinetic energy formula. Options C and D introduce incorrect fractional coefficients not present in derivation.
In inter-conversion of energy, the work done against the friction is:
A
\( f + h \)
B
\( f - h \)
C
\( fh \)
D
\( f/h \)
View Answer & Propolis Autopsy
Correct Key: Option CDiagnostic Explanation
Concept:
Work is fundamentally defined as the product of force and the displacement over which it acts.
Formula:
$$ W = F \times d $$
Solution:
Let the frictional force be \( f \).
Let the displacement (or height/distance fallen) be \( h \).
The magnitude of the work done to overcome this friction over that distance is the product of the two: \( W = f \times h = fh \).
Why other options are incorrect:
Adding, subtracting, or dividing force and distance violates dimensional analysis (you cannot add Newtons to meters, and N/m is a spring constant, not energy).
A common mistake is doing \( \frac{1}{2}m(v_f - v_i)^2 \), which yields \( 400(10)^2 = 40,000 \text{ J} = 40 \text{ kJ} \). Failing to convert Joules to kJ leads to magnitude errors.
When five times momentum of a body is equal to the kinetic energy of the same body then its velocity is equal to:
A
5 m/s
B
10 m/s
C
15 m/s
D
20 m/s
View Answer & Propolis Autopsy
Correct Key: Option BDiagnostic Explanation
Concept:
Set up an algebraic equation directly from the condition given in the word problem using standard formulas for momentum and kinetic energy.
Formula:
$$ 5p = K.E $$
Solution:
Substitute \( p = mv \) and \( K.E = \frac{1}{2}mv^2 \).
\( 5(mv) = \frac{1}{2}mv^2 \)
Cancel mass \( m \) and one \( v \) from both sides (assuming \( v \neq 0 \)).
\( 5 = \frac{1}{2}v \)
\( v = 5 \times 2 = 10 \text{ m/s} \).
Why other options are incorrect:
If you forget the \( 1/2 \) in the K.E formula, you get \( 5mv = mv^2 \Rightarrow v = 5 \text{ m/s} \) (Option A). Options C and D result from arbitrary arithmetic.
Kinetic energy is the energy possessed by a body due to its macroscopic motion.
Formula:
$$ K.E = \frac{1}{2}mv^2 $$
Solution:
The standard scalar definition of kinetic energy for a non-relativistic mass \( m \) moving at speed \( v \) is \( \frac{1}{2}mv^2 \).
Note: Option D (\( \frac{1}{2}m(\vec{v}\cdot\vec{v}) \)) is also mathematically true since the dot product of a velocity vector with itself gives the speed squared scalar, but Option B is the universal textbook representation.
Why other options are incorrect:
Option A is missing the square on momentum (it should be \( p^2/2m \)). Option C is dimensionally invalid.
A cyclist comes to a skidding stop in 10m. During this process, the opposing force on the cycle due to the road is 200N. How much work does the road do on the cycle?
A
-1800J
B
-2000J
C
2000J
D
1900J
View Answer & Propolis Autopsy
Correct Key: Option BDiagnostic Explanation
Concept:
Work done by a force opposing the direction of motion (like friction) is always negative because it removes kinetic energy from the system.
Formula:
$$ W = Fd \cos \theta $$
Solution:
Force of friction (\( F \)) = 200 N
Displacement (\( d \)) = 10 m
Since the force opposes motion, the angle \( \theta \) is \( 180^{\circ} \).
\( W = 200 \times 10 \times \cos(180^{\circ}) \)
\( W = 2000 \times (-1) = -2000 \text{ J} \)
Why other options are incorrect:
Option C (+2000J) implies the road pushed the bicycle forward, increasing its speed. Options A and D are calculation errors.
A body of mass 5kg is moving with a momentum of 10 Ns. a force of 0.2 N acts on it in the direction of motion of the body for 10s. The increase in its K.E is:
A
2.8 J
B
3.2 J
C
3.8 J
D
4.4 J
View Answer & Propolis Autopsy
Correct Key: Option DDiagnostic Explanation
Concept:
Use Newton's second law and kinematics to find the final velocity, then apply the work-energy theorem or calculate kinetic energy differences.
Formula:
$$ p = mv \quad ; \quad a = \frac{F}{m} \quad ; \quad v_f = v_i + at \quad ; \quad \Delta K.E = \frac{1}{2}m(v_f^2 - v_i^2) $$
The amount of work done is moving a body at certain point in a gravitational field to a position of zero potential such that the body is never accelerated is called:
A
Kinetic energy
B
Potential energy
C
Gravitational potential energy
D
Absolute potential energy
View Answer & Propolis Autopsy
Correct Key: Option DDiagnostic Explanation
Concept:
This is the formal definition of Absolute Gravitational Potential Energy.
Formula:
$$ U = -\frac{GMm}{r} $$
Solution:
"Gravitational potential energy" usually refers to relative changes (\( mgh \)) near the Earth's surface.
When moving a mass from a specific point all the way to infinity (where the gravitational field is strictly zero), the total work done without inducing acceleration defines the "Absolute" potential energy at that specific point.
Why other options are incorrect:
Kinetic energy relates to motion. Standard potential energy is relative (depends on an arbitrary reference frame like the floor or ground).
A field in which the work is done in a moving a body along a closed path is zero is called
A
Electric field
B
Conservative field
C
Electromagnetic field
D
Gravitational field
View Answer & Propolis Autopsy
Correct Key: Option BDiagnostic Explanation
Concept:
This is the literal definition of a conservative force field in classical mechanics.
Formula:
$$ \oint \vec{F} \cdot d\vec{r} = 0 $$
Solution:
If moving a mass around a closed loop results in net zero work done (energy spent getting there is fully recovered returning), the field governing it is broadly termed a "Conservative Field".
While Gravitational and Electrostatic fields are examples of conservative fields, the overarching generalized term for this property is "Conservative Field".
Why other options are incorrect:
While Options A and D are specific examples, Option B is the definitive, all-encompassing physical term the definition refers to.
An object in motion intrinsically possesses Kinetic Energy, which is a form of Mechanical Energy.
Formula:
$$ M.E = K.E + P.E $$
Solution:
Because the car is moving, it has velocity, giving it \( \frac{1}{2}mv^2 \).
Kinetic energy is categorized broadly as Mechanical energy.
Why other options are incorrect:
While a car produces heat and sound, and uses chemical energy (fuel), its pure state of gross physical motion is represented entirely by Mechanical energy.
Light body A and heavy body B have equal K. E of translation. Then
A
A has larger momentum than B
B
B has larger momentum than A
C
A and B have same momentum
D
None
View Answer & Propolis Autopsy
Correct Key: Option BDiagnostic Explanation
Concept:
Relate kinetic energy and momentum directly using algebra.
Formula:
$$ p = \sqrt{2mK.E} $$
Solution:
Kinetic Energy (K.E) is equal for both bodies.
The formula shows that momentum \( p \) is proportional to \( \sqrt{m} \).
Since body B is heavier (\( m_B > m_A \)), it will have a larger momentum (\( p_B > p_A \)).
Why other options are incorrect:
People confuse the velocity relationships. A light body moves much faster to have the same KE, but momentum heavily favors the mass component when KE is held constant.
A 1.75 m heighted weight-lifter raises weights with a mass of 50 kg to a height of .5m above his head. How much work is being done by him? (\( g=10 \text{ ms}^{-2} \))
A
2125J
B
2500 J
C
50J
D
1125 J
View Answer & Propolis Autopsy
Correct Key: Option DDiagnostic Explanation
Concept:
Work done against gravity is the product of weight (mg) and the total vertical displacement from the ground.
Formula:
$$ W = mgh $$
Solution:
Mass (\( m \)) = 50 kg
Total vertical height from ground (\( h \)) = Lifter's height + extra height = 1.75 + 0.5 = 2.25 m
Which of the following force is non-conservative force?
A
Frictional force
B
Gravitation force
C
Electric force
D
Elastic spring force
View Answer & Propolis Autopsy
Correct Key: Option ADiagnostic Explanation
Concept:
A non-conservative force dissipates mechanical energy (often into heat) and the work done by it depends entirely on the path taken, not just initial and final positions.
Formula:
$$ W_{\text{closed}} \neq 0 $$
Solution:
Frictional force opposes motion constantly. If you move an object in a closed loop, friction does negative work the entire time, dissipating energy.
Therefore, friction is path-dependent and non-conservative.
Why other options are incorrect:
Gravitational, electric, and spring forces are conservative. Energy stored against them can be perfectly recovered as kinetic energy.
An automobile is moving forwards with uniform velocity due to the force exerted by its engine. If that force is double with the velocity remaining constant what happens to its total power?
A
It does not change
B
It is halved
C
It is squared
D
It is doubled
View Answer & Propolis Autopsy
Correct Key: Option DDiagnostic Explanation
Concept:
Power output is directly proportional to both the force exerted and the velocity at which the object moves.
Formula:
$$ P = F \times v $$
Solution:
Initial Power \( P = Fv \)
New Force = \( 2F \)
Velocity remains \( v \)
New Power \( P' = (2F)v = 2(Fv) = 2P \)
Thus, the power is doubled.
Why other options are incorrect:
Since the relationship is strictly linear, doubling one variable (while keeping the other constant) doubles the product. It does not square it or halve it.
Which of the following statement shows that no work is done?
A
Pushing a car to start it moving
B
Lifting the weights
C
Writing an essay on a page
D
The moon orbiting the earths
View Answer & Propolis Autopsy
Correct Key: Option DDiagnostic Explanation
Concept:
In physics, mechanical work is zero if the force acts perpendicularly to the direction of motion at all times.
Formula:
$$ W = Fd \cos(90^{\circ}) = 0 $$
Solution:
For a moon orbiting Earth in a circular path, the gravitational force provides centripetal force, pointing towards Earth's center.
The velocity (and instantaneous displacement) is tangential, forming a 90° angle with the force.
Therefore, \( \cos(90^{\circ}) = 0 \), so no work is done.
Why other options are incorrect:
Pushing a car and lifting weights involve forces applied in the same direction as displacement, doing positive work. Writing involves frictional forces and micro-displacements, doing work.
A stone of mass 2.0 kg is dropped from a rest position 5.0m above the ground. What is its velocity at a height of 3.0m above the ground?
A
12.5m/s
B
9.3m/s
C
6.3m/s
D
16.0m/s
View Answer & Propolis Autopsy
Correct Key: Option CDiagnostic Explanation
Concept:
Using the Law of Conservation of Energy, the loss in potential energy equals the gain in kinetic energy. The velocity depends only on the distance it has fallen, not the mass.
Formula:
$$ v = \sqrt{2g\Delta h} $$
Solution:
Initial height = 5.0 m
Final height = 3.0 m
Distance fallen (\( \Delta h \)) = 5.0 - 3.0 = 2.0 m
A man has a mass of 80 kg. He ties himself to one end of rope which passes over a single fixed pulley. He pulls on the other end of the rope to lift himself up at an average speed of \( 50 \text{ cms}^{-1} \). What is the average useful power at which he is working?
A
40W
B
0.39kW
C
4.0kW
D
39kW
View Answer & Propolis Autopsy
Correct Key: Option BDiagnostic Explanation
Concept:
Power can be calculated as the product of the upward force applied (to overcome gravity) and the constant velocity of the mass.
Convert to kilowatts: \( \frac{392}{1000} = 0.392 \text{ kW} \)
Why other options are incorrect:
If \( g=10 \) is used, \( P = 800 \times 0.5 = 400 \text{ W} = 0.4 \text{ kW} \). The closest exact answer using 9.8 is 0.39 kW. Failing to convert cm/s to m/s results in large values like 39 kW.
A man carries a 1 kg body 10m horizontally on a level ground. The work done by the man is:
A
10 J
B
1 J
C
0 J
D
5 J
View Answer & Propolis Autopsy
Correct Key: Option CDiagnostic Explanation
Concept:
Work is calculated as the dot product of force and displacement. When moving horizontally at constant speed, the applied lifting force is upward (against gravity), while the displacement is horizontal.
Formula:
$$ W = Fd \cos \theta $$
Solution:
The force holding the object up is vertical.
The displacement is horizontal.
The angle \( \theta \) between force and displacement is 90°.
\( \cos(90^{\circ}) = 0 \)
\( W = Fd(0) = 0 \text{ J} \)
Why other options are incorrect:
Multiplying mass and distance (1 x 10) yields 10, which incorrectly assumes the force and displacement are in the same direction.
The net work done by a conservative force in moving an object along a closed path (returning to the exact starting position) is always zero.
Formula:
$$ \oint \vec{F} \cdot d\vec{r} = 0 $$
Solution:
The figure represents a cyclic process forming a closed loop on a Force-Displacement graph.
In a closed path under a conservative field, the positive work done expanding is exactly canceled by the negative work done returning to the origin.
Thus, the net work is 0 Nm.
Why other options are incorrect:
Calculating the geometric area inside the loop gives the work done during one specific cycle segment, but the total net displacement from start to finish is zero, yielding zero net work.
If mass 'm' is dropped from height 'h' vertically, f is the force of friction during downward motion and 'v' is the velocity at bottom, following equation will be hold:
A
\( \frac{1}{2} mv^2 = mgh + fh \)
B
\( fh = mgh + \frac{1}{2} mv^2 \)
C
\( mgh = \frac{1}{2} mv^2 - fh \)
D
\( mgh = \frac{1}{2} mv^2 + fh \)
View Answer & Propolis Autopsy
Correct Key: Option DDiagnostic Explanation
Concept:
According to the Conservation of Energy in the presence of non-conservative forces, the initial potential energy is converted into kinetic energy plus the work done against friction.
Initial energy at height \( h \) is entirely Potential Energy: \( P.E = mgh \)
Final energy at the bottom is Kinetic Energy: \( K.E = \frac{1}{2}mv^2 \)
Work done against air friction over distance \( h \): \( W = fh \)
Equating them: \( mgh = \frac{1}{2}mv^2 + fh \)
Why other options are incorrect:
Option A implies energy is created. Option B implies friction work equals total energy. Option C incorrectly subtracts friction work instead of adding it to the final state energies.
Work depends on the cosine of the angle between the applied force vector and the displacement vector.
Formula:
$$ W = Fd \cos \theta $$
Solution:
The cosine function reaches its maximum positive value when the angle \( \theta = 0^{\circ} \).
\( \cos(0^{\circ}) = 1 \)
Therefore, the work done is \( W = Fd(1) = Fd \), which is the maximum possible work.
Why other options are incorrect:
At 90°, work is zero. At 30° and 45°, the cosine values are \( \frac{\sqrt{3}}{2} \) and \( \frac{1}{\sqrt{2}} \) respectively, which yield fractions of the maximum work.
Work is a scalar quantity. The positive (+) or negative (-) signs merely indicate whether energy is being transferred to the system (positive) or from the system (negative, like work done against friction). "Greater work" refers to the largest absolute magnitude of energy transferred.
Formula:
$$ |W| = \text{Magnitude of energy transfer} $$
Solution:
Compare the absolute magnitudes of the given options:
|+100| = 100 J
|-1000| = 1000 J
|-100| = 100 J
|+200| = 200 J
1000 J is the greatest magnitude of energy transferred.
Why other options are incorrect:
Students often mistake the (-) sign for a mathematical algebraic value (where -1000 is smaller than +100). In physics, a negative work of 1000 J means 1000 J of energy was removed, which is a "greater" amount of work than adding 100 J.
The figure shows the force distance curve of a body moving along a straight line. The work done by the force:
A
10 J
B
30 J
C
20 J
D
40 J
View Answer & Propolis Autopsy
Correct Key: Option ADiagnostic Explanation
Concept:
The work done by a variable force is equal to the net area under the Force-Displacement (F-d) graph. Areas above the x-axis represent positive work, and areas below represent negative work.
Formula:
$$ W = \sum \text{Area} $$
Solution:
Based on standard graph variations of this problem: The graph consists of three rectangular segments.
Segment 1: Force = +10 N, Displacement = 1 m. Area = \( 10 \times 1 = +10 \text{ J} \)
Segment 2: Force = -10 N, Displacement = 1 m. Area = \( -10 \times 1 = -10 \text{ J} \)
Segment 3: Force = +10 N, Displacement = 1 m. Area = \( 10 \times 1 = +10 \text{ J} \)
Net Work = \( (+10) + (-10) + (+10) = 10 \text{ J} \)
Why other options are incorrect:
Ignoring the negative sign for areas below the x-axis results in 30 J (Option B). Only looking at two positive segments gives 20 J (Option C).
Using the slant length of the staircase (20m) instead of vertical height yields \( 60 \times 10 \times 20 = 12000 \text{ J} \) or 12 KJ (Option A), which is incorrect because gravity only acts vertically.
When a force retards the motion of a body the work done is:
A
Zero
B
Negative
C
Positive
D
Positive or negative depending upon the magnitude of force and displacement
View Answer & Propolis Autopsy
Correct Key: Option BDiagnostic Explanation
Concept:
A retarding force is one that opposes the direction of motion (e.g., friction or braking force). In this case, the angle between force and displacement is 180°.
Formula:
$$ W = Fd \cos \theta $$
Solution:
Angle \( \theta = 180^{\circ} \)
\( \cos(180^{\circ}) = -1 \)
\( W = Fd(-1) = -Fd \)
The work done is fundamentally negative because energy is being removed from the moving body.
Why other options are incorrect:
Positive work implies the force aids motion (speeding the object up). Zero work implies force acts perpendicularly. Option D is incorrect because retardation strictly implies opposition.
An engine pumps out 40 kg of water in one second. The water comes out vertically upwards with a velocity of \( 3 \text{ ms}^{-1} \), the power of engine in kilowatt is:
A
1.2 kW
B
12 kW
C
120 kW
D
1200 kW
View Answer & Propolis Autopsy
Correct Key: Option ADiagnostic Explanation
Concept:
Power can be calculated using the force exerted and the velocity at which the mass moves. Since water is pumped vertically, the force must overcome gravity.
Formula:
$$ P = F \cdot v = (mg) \cdot v $$
Solution:
Mass per second (\( m/t \)) = 40 kg/s
Velocity (\( v \)) = \( 3 \text{ m/s} \)
Assume \( g = 10 \text{ m/s}^2 \) for standard ETEA simplicity.
If you multiply the direct ratios without flipping time (4/5 * 5/4), you get 1 (Option A). If you flip weight instead of time, you get 25/16 (Option C).
A 6.0 kg block is released from rest 80m above the ground. When it has fallen 60m its kinetic energy is approximately:
A
4800 J
B
3500 J
C
1200 J
D
120 J
View Answer & Propolis Autopsy
Correct Key: Option BDiagnostic Explanation
Concept:
According to the Law of Conservation of Energy, the loss in gravitational potential energy equals the gain in kinetic energy for a freely falling body.
Formula:
$$ \text{Loss in P.E} = \text{Gain in K.E} = mg\Delta h $$
Solution:
Mass (\( m \)) = 6.0 kg
Distance fallen (\( \Delta h \)) = 60 m (Note: we use the distance it has fallen, not its height from the ground).
Using the remaining height of 20m instead of the fallen distance gives \( 6 \times 9.8 \times 20 \approx 1200 \text{ J} \) (Option C). Using the total height 80m gives \( \approx 4800 \text{ J} \) (Option A).
Which one of the following is a non-conservative force?
A
Electric force
B
Gravitational force
C
Elastic spring force
D
Frictional force
View Answer & Propolis Autopsy
Correct Key: Option DDiagnostic Explanation
Concept:
A non-conservative force is a force for which the work done depends on the path taken. The mechanical energy of a system is not conserved when non-conservative forces act (it dissipates as heat or sound).
Formula:
$$ W_{\text{closed path}} \neq 0 $$
Solution:
Electric, gravitational, and elastic spring forces are all conservative forces because work done by them over a closed loop is zero.
Frictional force is path-dependent and dissipates energy as heat, making it a classic non-conservative force.
Why other options are incorrect:
Options A, B, and C are fundamental conservative fields where potential energy can be perfectly defined and recovered.
The gravitational potential energy per unit mass is called:
A
Gravitational potential
B
Absolute P.E
C
P.E
D
Potential hill
View Answer & Propolis Autopsy
Correct Key: Option ADiagnostic Explanation
Concept:
Gravitational potential at a point in a gravitational field is defined as the work done per unit mass to bring a test mass from infinity to that point.
Formula:
$$ V_g = \frac{U}{m} = \frac{mgh}{m} = gh $$
Solution:
Potential Energy (\( U \)) is energy associated with a specific mass \( m \).
When we divide this energy by the mass (\( U/m \)), we get a property of the field itself, independent of the test mass. This is called Gravitational Potential.
Why other options are incorrect:
Absolute P.E and regular P.E are measures of total energy (Joules), not energy per unit mass (Joules/kg). A potential hill is a visual analogy, not the physical quantity.
Option D represents only the final kinetic energy. Options A and C result from arbitrary arithmetic (like taking averages) rather than applying the direct difference.
Subtracting the coordinates incorrectly or failing to apply the dot product rule (e.g., multiplying mismatched components) leads to incorrect negative or larger values like -10, -18, or +18.
Work done on a body equals change in its ____ energy.
A
Total
B
Kinetic
C
Potential
D
All of these
View Answer & Propolis Autopsy
Correct Key: Option DDiagnostic Explanation
Concept:
The generalized Work-Energy Principle states that the net work done on a system can manifest as a change in kinetic energy, potential energy, or total mechanical energy, depending on the forces acting (conservative vs. non-conservative).
Formula:
$$ W_{\text{net}} = \Delta K.E + \Delta P.E $$
Solution:
Work done on a freely moving body changes its Kinetic Energy (Work-Kinetic Energy Theorem).
Work done against a conservative force (like gravity or a spring) changes its Potential Energy.
Work done on a mass-spring system can change the Total Energy.
Therefore, work can equal the change in any of these energies depending on the exact conditions.
Why other options are incorrect:
While work equals change in kinetic energy specifically for the net force, and change in potential energy for conservative forces, stating only one of them is incomplete in a general sense.
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