The displacement x of a body of mass 1 kg on smooth horizontal surface as a function of time t is given by x = t 3 3 (where x is in metres and t is in seconds). Find the work done by the external agent for the first second.

Finding the instantaneous velocity and applying work energy theorem holds good here. Work energy theorem states that the work done by all the external forces equals in the change in kinetic energy of the body. x = t 3 3 ⇒ v = d x d t = t 2 The kinetic energy K E = 1 2 m v 2 = 1 2 m t 4 Initial kinetic energy K E i = 0 The final kinetic energy K E f = 1 2 × 1 × 1 2 ⇒ K E f = 1 2 = 0 . 5 J Therefore, the work done by the external agent W = K E f - K E i = 0 . 5 - 0 ⇒ W = 0 . 5 J

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Beta Question Bank for Medical: Physics>Chapter 8 - Work, Energy and Power>BEGINNER'S BOX - 2

Embibe Experts Solutions for Chapter: Work, Energy and Power, Exercise 2: BEGINNER'S BOX - 2

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Embibe Experts Physics Solutions for Exercise - Embibe Experts Solutions for Chapter: Work, Energy and Power, Exercise 2: BEGINNER'S BOX - 2

Attempt the practice questions on Chapter 8: Work, Energy and Power, Exercise 2: BEGINNER'S BOX - 2 with hints and solutions to strengthen your understanding. Beta Question Bank for Medical: Physics solutions are prepared by Experienced Embibe Experts.

Questions from Embibe Experts Solutions for Chapter: Work, Energy and Power, Exercise 2: BEGINNER'S BOX - 2 with Hints & Solutions

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Beta Question Bank for Medical: Physics>Chapter 8 - Work, Energy and Power>BEGINNER'S BOX - 2>Q 1

The kinetic energy of a body is numerically equal to thrice the momentum of the body. Find the velocity of the body.

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Beta Question Bank for Medical: Physics>Chapter 8 - Work, Energy and Power>BEGINNER'S BOX - 2>Q 2

A body of mass $10 kg$ is released from the top of a tower which acquires a velocity of $10 m s^{- 1}$ after falling through the distance of $20 m$ . Calculate the work done by the drag force of the air on the body? (take $g = 10 m s^{- 2}$ )

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Beta Question Bank for Medical: Physics>Chapter 8 - Work, Energy and Power>BEGINNER'S BOX - 2>Q 3

The displacement $x$ of a body of mass $1 kg$ on smooth horizontal surface as a function of time $t$ is given by $x =$ $\frac{t^{3}}{3}$ (where $x$ is in metres and $t$ is in seconds). Find the work done by the external agent for the first second.