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JEE PHYSICS MCQ: WORK-ENERGY-POWER
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A force acts on a 2 kg object, so that its position is given as a function of time as . What is the work done by this force in first 5 seconds?
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Correct option: B
900 J
Option B (900 J) is correct because the work done is equal to the change in kinetic energy, which can be calculated from the velocity at t=0 and t=5 seconds.
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Assertion:When a particle moves in a circle with a uniform speed its acceleration is constant but the velocity changes.
Reason: Angular displacement is not an axial vector. A body of mass thrown vertically upwards attains a maximum height . As what height will its kinetic energy be of its initial value? Assertion: Centripetal force does no work.
Reason: Force and displacement are perpendicular to each other. Statement I: Both stretched and compressed springs possess potential energy.
Statement II: Work done against restoring force is stored as potential energy. A bullet of mass moving at a speed hits a ball of mass kept at rest. A small part having mass breaks from the ball and sticks to the bullet. The remaining ball is found to move at a speed in the direction of the bullet. Find the velocity of the bullet after the collision.
Reason: Angular displacement is not an axial vector. A body of mass thrown vertically upwards attains a maximum height . As what height will its kinetic energy be of its initial value? Assertion: Centripetal force does no work.
Reason: Force and displacement are perpendicular to each other. Statement I: Both stretched and compressed springs possess potential energy.
Statement II: Work done against restoring force is stored as potential energy. A bullet of mass moving at a speed hits a ball of mass kept at rest. A small part having mass breaks from the ball and sticks to the bullet. The remaining ball is found to move at a speed in the direction of the bullet. Find the velocity of the bullet after the collision.