A particle of mass 0 . 5 kg travels in a straight line with velocity v = a x 3 2 where a = 5 m - 1 2 s - 1 . What is the work done by the net force during its displacement from x = 0 to x = 2 m ?

Work energy theorem states, work done by net forces is equal to change is the kinetic energy of the particle. Kinetic energy K E = 1 2 m v 2 K E = 1 2 m a x 3 2 2 , K E i = 1 2 × 0 . 5 a 2 × 0 ⇒ K E i = 0 , K E f = 1 2 × 0 . 5 a 2 × 2 3 ⇒ K E f = 2 a 2 ⇒ K E f = 50 J , The work done by net force W = K E f - K E i = 50 - 0 ⇒ W = 50 J

A particle moves from position r → 1 = 3 i ^ + 2 j ^ - 6 k ^ m to position r → 2 = 14 i ^ + 13 j ^ + 9 k ^ m under the action of force 4 i ^ + j ^ + 3 k ^ N . Find the work done.

The work done by the force is the scalar product of the force vector and displacement vector. Work done is a scalar quantity W = F → · s → . The displacement of the particle, s → = r → 2 - r → 1 s → = 14 - 3 i ^ + 13 - 2 j ^ + 9 + 6 k ^ ⇒ s → = 11 i ^ + 11 j ^ + 15 k ^ . The work done, W = F → · s → = 4 i ^ + j ^ + 3 k ^ · 11 i ^ + 11 j ^ + 15 k ^ ⇒ W = 100 J .

F → = K ( x i ^ + y j ^ ) is working on a particle that moves from ( 0 , 0 ) to ( a , 0 ) along x - axis and ( a , 0 ) to ( a , a ) along y -axis. Find the work done by the force.

Work done by the variable force W = ∫ F d x The particle starts from ( 0 , 0 ) and reaches ( a , a ) The work done by conservative force does not depend upon the path, it depends upon initial and final conditions. ⇒ W = K ∫ 0 , 0 a , a x ( d y ) + y ( d x ) ⇒ W = K ∫ 0 , 0 a , a d ( x y ) ⇒ W = K x y 0 , 0 a , a ⇒ W = K a 2

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Assertion : For an isolated system mechanical energy is conserved.
Reason : Kinetic frictional force is absent for an isolated system.

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Important Questions on Work, Energy and Power

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

In Fig. (i), the man walks 2 m carrying a mass of 15 kg on his hand. In Fig. (ii), he walks the same distance pulling the rope behind him. The rope goes over a pulley, and a mass of 15 kg hangs at its other end. In which case is the work done by gravity is greater ? Question Image

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Beta Question Bank for Medical: Physics>Chapter 8 - Work, Energy and Power>BEGINNER'S BOX - 1>Q 4

A force

F = (10 + 0.5 x) N

acts on a particle in

x

-direction, where

x

is in

metres

. Find the work done by this force during a displacement from

x = 0

x = 2

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Beta Question Bank for Medical: Physics>Chapter 8 - Work, Energy and Power>BEGINNER'S BOX - 1>Q 5

A particle of mass

0.5 kg

travels in a straight line with velocity

v = a x^{\frac{3}{2}}

where

a = 5 m^{- \frac{1}{2}} s^{- 1}

. What is the work done by the net force during its displacement from

x = 0

x = 2 m

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Beta Question Bank for Medical: Physics>Chapter 8 - Work, Energy and Power>BEGINNER'S BOX - 1>Q 6

A particle moves from position

{\vec{r}}_{1} = (3 \hat{i} + 2 \hat{j} - 6 \hat{k}) m

to position

{\vec{r}}_{2} = (14 \hat{i} + 13 \hat{j} + 9 \hat{k}) m

under the action of force

(4 \hat{i} + \hat{j} + 3 \hat{k}) N

. Find the work done.

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Beta Question Bank for Medical: Physics>Chapter 8 - Work, Energy and Power>BEGINNER'S BOX - 1>Q 7

\vec{F} = K (x \hat{i} + y \hat{j})

is working on a particle that moves from

(0, 0)

(a, 0)

along

x -

axis and

(a, 0)

(a, a)

along

y

-axis. Find the work done by the force.

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Beta Question Bank for Medical: Physics>Chapter 8 - Work, Energy and Power>BEGINNER'S BOX - 1>Q 9

A horizontal force $F$ is used to pull a box placed on a floor. Variation in the force with position coordinate $x$ measured along the floor is shown in the graph.

Question Image
(a) Calculate the work done by the force in moving the box from $x = 0$ to $x = 10 m$ .
(b) Calculate the work done by the force in moving the box from $x = 10 m$ to $x = 15 m$ .
(c) Calculate the work done by the force in moving the box from $x = 0$ to $x = 15 m$ .

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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}

)

Assertion : For an isolated system mechanical energy is conserved. Reason : Kinetic frictional force is absent for an isolated system.

Important Questions on Work, Energy and Power

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Assertion : For an isolated system mechanical energy is conserved.
Reason : Kinetic frictional force is absent for an isolated system.