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Name: Work and Force Physics mcqs test
Subject: Physics
Topic: Work and Force
Questions: 8 MCQ
Time Allowed: 18 Minutes
Important for: 11th & 12th School students, Engineering and Medical Entrance exams.
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 Question 1 of 8
1. Question
1 pointsWhen a body moves with a constant velocity, then
CorrectWe know that work done (W) = Force × Distance
= (Mass × Acceleration) × Distance.
Since the body moves with a constant speed, therefore acceleration of the body is zero. And the force is also zero. Or in other words, no work is done on it.IncorrectWe know that work done (W) = Force × Distance
= (Mass × Acceleration) × Distance.
Since the body moves with a constant speed, therefore acceleration of the body is zero. And the force is also zero. Or in other words, no work is done on it.UnattemptedWe know that work done (W) = Force × Distance
= (Mass × Acceleration) × Distance.
Since the body moves with a constant speed, therefore acceleration of the body is zero. And the force is also zero. Or in other words, no work is done on it.  Question 2 of 8
2. Question
1 pointsThe work done by a force is given by the dot product of force and displacement. If work done is zero, then which of the following statement is correct regarding force and displacement?
CorrectWe have: Work done (W) = 0
We know that work done (W) = F s cosθ or F s cosθ = 0
Since F and s are not equal to zero, therefore θ=90°. Or in the other words, force and displacement are at right angle to each other. Thus option ‘Both are at right angle to each other’ is correct.IncorrectWe have: Work done (W) = 0
We know that work done (W) = F s cosθ or F s cosθ = 0
Since F and s are not equal to zero, therefore θ=90°. Or in the other words, force and displacement are at right angle to each other. Thus option ‘Both are at right angle to each other’ is correct.UnattemptedWe have: Work done (W) = 0
We know that work done (W) = F s cosθ or F s cosθ = 0
Since F and s are not equal to zero, therefore θ=90°. Or in the other words, force and displacement are at right angle to each other. Thus option ‘Both are at right angle to each other’ is correct.  Question 3 of 8
3. Question
1 pointsA particle is acted upon by a constant force perpendicular to the direction of motion of the particle, which of the following statement about the particle is correct?
CorrectWe know that as the constant force acting on the particle is perpendicular to its direction of motion, therefore work done by the force is zero.
Therefore kinetic energy of the particle is constant. Thus option ‘Its kinetic energy is constant’ is correct.IncorrectWe know that as the constant force acting on the particle is perpendicular to its direction of motion, therefore work done by the force is zero.
Therefore kinetic energy of the particle is constant. Thus option ‘Its kinetic energy is constant’ is correct.UnattemptedWe know that as the constant force acting on the particle is perpendicular to its direction of motion, therefore work done by the force is zero.
Therefore kinetic energy of the particle is constant. Thus option ‘Its kinetic energy is constant’ is correct.  Question 4 of 8
4. Question
1 pointsA body moves a distance of 10m along a straight line under the action of a 5N force. If work done is 25J, then angle between the force and the direction of motion of the body is
CorrectWe have: Distance covered by body (s) = 10m; Force (F) = 5N and work done (W) = 25J.
We know that work done (W) = F s cosθ or cosθ = W/Fs = 25 / (5×10) = ½
or θ = 60° (where θ = Angle between the force and the direction of motion).IncorrectWe have: Distance covered by body (s) = 10m; Force (F) = 5N and work done (W) = 25J.
We know that work done (W) = F s cosθ or cosθ = W/Fs = 25 / (5×10) = ½
or θ = 60° (where θ = Angle between the force and the direction of motion).UnattemptedWe have: Distance covered by body (s) = 10m; Force (F) = 5N and work done (W) = 25J.
We know that work done (W) = F s cosθ or cosθ = W/Fs = 25 / (5×10) = ½
or θ = 60° (where θ = Angle between the force and the direction of motion).  Question 5 of 8
5. Question
1 pointsA body of mass 50kg slides over a horizontal distance of 1m. If coefficient of friction between their surface is 0.2, then work done against friction is (Take g= 9.8 ms2)
CorrectGiven: Mass of body (m) = 50kg; Displacement (s) = 1m; Coefficient of friction (μ) = 0.2 acceleration due to gravity (g) = 9.8 ms^{2}.
We know that normal reaction (R) = mg = 50 × 9.8 = 490 N.
And frictional force (F) = μR = 0.2 × 490 = 98 N.
Therefore work done against friction (W) = Fs = 98 × 1 = 98 J.IncorrectGiven: Mass of body (m) = 50kg; Displacement (s) = 1m; Coefficient of friction (μ) = 0.2 acceleration due to gravity (g) = 9.8 ms^{2}.
We know that normal reaction (R) = mg = 50 × 9.8 = 490 N.
And frictional force (F) = μR = 0.2 × 490 = 98 N.
Therefore work done against friction (W) = Fs = 98 × 1 = 98 J.UnattemptedGiven: Mass of body (m) = 50kg; Displacement (s) = 1m; Coefficient of friction (μ) = 0.2 acceleration due to gravity (g) = 9.8 ms^{2}.
We know that normal reaction (R) = mg = 50 × 9.8 = 490 N.
And frictional force (F) = μR = 0.2 × 490 = 98 N.
Therefore work done against friction (W) = Fs = 98 × 1 = 98 J.  Question 6 of 8
6. Question
1 pointsA block of mass 1kg slides down a rough inclined plane of inclination 60° starting from its top. If coefficient of kinetic friction is 0.5 and length of the plane is 1m, then work done against friction is
CorrectGiven: Mass of the block (m) = 1 kg; Inclination of place (θ) = 60°; Coefficient of friction (μ) = 0.5 and length of the plane (s) = 1m.
We know that normal reaction on the plane (R) = m g cosθ = 1 × 9.8 × cos60° = 1×9.8×0.5 = 4.9 N.
And frictional force (F) = μR = 0.5×4.9 = 2.45N.
Therefore work done against friction (W) = F s = 2.45 × 1 = 2.45 J.IncorrectGiven: Mass of the block (m) = 1 kg; Inclination of place (θ) = 60°; Coefficient of friction (μ) = 0.5 and length of the plane (s) = 1m.
We know that normal reaction on the plane (R) = m g cosθ = 1 × 9.8 × cos60° = 1×9.8×0.5 = 4.9 N.
And frictional force (F) = μR = 0.5×4.9 = 2.45N.
Therefore work done against friction (W) = F s = 2.45 × 1 = 2.45 J.UnattemptedGiven: Mass of the block (m) = 1 kg; Inclination of place (θ) = 60°; Coefficient of friction (μ) = 0.5 and length of the plane (s) = 1m.
We know that normal reaction on the plane (R) = m g cosθ = 1 × 9.8 × cos60° = 1×9.8×0.5 = 4.9 N.
And frictional force (F) = μR = 0.5×4.9 = 2.45N.
Therefore work done against friction (W) = F s = 2.45 × 1 = 2.45 J.  Question 7 of 8
7. Question
1 pointsOnefourth chain is hanging down from a table. The work done to bring the hanging part of the chain on the table is (Masss of chain = M and length = L)
CorrectWe know that Mass of chain = M; Total length of the chain = L and the length of the chain hanging down from the table (l) = L/4.
We know that height through which the centre of mass of the hanging part of the chain is to be lifted (h) = l/2 = (L/4)/2 = L/8.
And mass of the hanging part of the chain (m) = M × (l/L) = M × (L/4)/L = M/4.
Therefore work done to bring the hanging chain on the table = Force × Distance
= (mg)×h = (M/4 ×g) × (L/8) = MgL/32.IncorrectWe know that Mass of chain = M; Total length of the chain = L and the length of the chain hanging down from the table (l) = L/4.
We know that height through which the centre of mass of the hanging part of the chain is to be lifted (h) = l/2 = (L/4)/2 = L/8.
And mass of the hanging part of the chain (m) = M × (l/L) = M × (L/4)/L = M/4.
Therefore work done to bring the hanging chain on the table = Force × Distance
= (mg)×h = (M/4 ×g) × (L/8) = MgL/32.UnattemptedWe know that Mass of chain = M; Total length of the chain = L and the length of the chain hanging down from the table (l) = L/4.
We know that height through which the centre of mass of the hanging part of the chain is to be lifted (h) = l/2 = (L/4)/2 = L/8.
And mass of the hanging part of the chain (m) = M × (l/L) = M × (L/4)/L = M/4.
Therefore work done to bring the hanging chain on the table = Force × Distance
= (mg)×h = (M/4 ×g) × (L/8) = MgL/32.  Question 8 of 8
8. Question
1 pointsA uniform chain of length 2m is kept on a table; such that a length of 60cm hangs freely form the edge of the table. The total mass of the chain is 4kg. What is the work done in pulling the entire chain on the table?
CorrectGiven: Total length of chain (L) = 2m;
Length of the chain hanging down from the table (l) = 60cm = 0.6m and total mass of the chain (M) = 4kg.
We know that height through which the centre of mass of the hanging part of the chain is to be lifted (h) = l/2 = 0.6/2 = 0.3m.
And mass of the hanging part of the chain (m) = M × (l/L) = 4×(0.6/2) = 1.2kg.
Therefore work done in pulling the entire chain on the table = Force × Distance = (mg) × h = (1.2×10)×0.3 = 3.6 Nm = 3.6 J.IncorrectGiven: Total length of chain (L) = 2m;
Length of the chain hanging down from the table (l) = 60cm = 0.6m and total mass of the chain (M) = 4kg.
We know that height through which the centre of mass of the hanging part of the chain is to be lifted (h) = l/2 = 0.6/2 = 0.3m.
And mass of the hanging part of the chain (m) = M × (l/L) = 4×(0.6/2) = 1.2kg.
Therefore work done in pulling the entire chain on the table = Force × Distance = (mg) × h = (1.2×10)×0.3 = 3.6 Nm = 3.6 J.UnattemptedGiven: Total length of chain (L) = 2m;
Length of the chain hanging down from the table (l) = 60cm = 0.6m and total mass of the chain (M) = 4kg.
We know that height through which the centre of mass of the hanging part of the chain is to be lifted (h) = l/2 = 0.6/2 = 0.3m.
And mass of the hanging part of the chain (m) = M × (l/L) = 4×(0.6/2) = 1.2kg.
Therefore work done in pulling the entire chain on the table = Force × Distance = (mg) × h = (1.2×10)×0.3 = 3.6 Nm = 3.6 J.