When a body falls freely under gravity, then the work done by the gravity is?
A.
Positive
B.
Negative
C.
Zero
D.
Infinity
Answer: Option A
Explanation:
If a force acting on a body has a component in the direction of displacement, then the work done by the force is positive. Hence when a body falls freely under the influence of gravity the work done by the gravity is positive.
When a body slides against a rough horizontal surface, the work done by friction is?
A.
Positive
B.
Zero
C.
Negative
D.
Constant
Answer: Option C
Explanation:
If a force acting on a body has a component in the opposite direction of displacement, the work done is negative, when a body slides against a rough horizontal surface, its displacement is opposite to that of the force of friction. He works done by the friction is negative.
For a body moving in a circular path, the work done by the centripetal force is?
A.
Negative
B.
Positive
C.
Constant
D.
Zero
Answer: Option D
Explanation:
For a body moving in a circular path, the centripetal force and the displacement are perpendicular to each other. So the work done by the centripetal force is zero.
A gardener pushes a lawn roller through a distance of 20m. If he applies a force of 20kg weight in a direction inclined at 60° to the ground, find the work done by him. (g=9.8m/s2)?
A.
400J
B.
1960J
C.
250J
D.
2514J
Answer: Option B
Explanation:
F = 20kg wt = 20×9.8N s = 20m θ = 60° W = Fscosθ = 20×9.8×20×cosâ¡60° W = 1960J.
A person is holding a bucket by applying a force of 10N. He moves a horizontal distance of 5m and then climbs up a vertical distance of 10m. Find the total work done by him?
A.
50J
B.
150J
C.
100J
D.
200J
Answer: Option C
Explanation:
F = 10N, s = 5m, θ = 90° Work done, W1=Fscosθ = 10×5×cos90° = 0 For vertical motion, the angle between force and displacement is 0°. Here, F = 10N, s = 10m, θ=0° Work done, W2=10×10×cos0 = 100J Total work done = W1+W2 = 100J.
