Question

A body of mass 2 kg initially at rest moves under the action of an applied horizontal force of 7 N on a table with coefficient of kinetic friction = 0.1. Compute the (a) work done by the applied force in 10 s, (b) work done by friction in 10 s, (c) work done by the net force on the body in 10 s, (d) change in kinetic energy of the body in 10 s, and interpret your results.

Answer 1

Given that the mass of the body is 2 kg, The magnitude of the horizontal force is 7 N and the magnitude of coefficient of kinetic friction is 0.1.

a)

The formula to calculate the frictional force is,

F f =μmg

Here, the mass of the body is m, the magnitude of the kinetic friction is μ, and the gravitational acceleration is g.

Substitute the values in the above equation.

F f =( 0.1 )( 2 )( 9.8 ) =1.96 N

The net force that is applied on the body is,

F n =F− F f

Here, the applied horizontal force is F and the kinetic frictional force is F f .

Substitute the values in the above equation.

F n =7−1.96 =5.04 N

The formula to calculate the acceleration is,

a= F n m

Substitute the values in the above equation.

a= 5.04 2 =2.52 m/ s 2

The formula to calculate the distance travelled by the body due to the net force is,

s= 1 2 a t 2

Here, the acceleration of the body is a and the time for which the body is accelerated is t.

Substitute the values in the above equation.

s= 1 2 ( 2.52 ) ( 10 ) 2 =126 m

The formula to calculate the work done by the applied force is,

W 1 =Fs

Substitute the value in the above equation.

W 1 =( 7 )( 126 ) =882 J

Thus, the work done by the applied force in 10 s is 882 J.

b)

The formula to calculate the work done is,

W 2 = F f scosθ

Here, the kinetic force of friction is F f , the displacement of the body is s and the angle between the force and the displacement is θ.

Substitute the value in the above equation.

W 2 =( 1.96 )( 126 )cos180° ≈−247 J

Thus, the work done by the friction force in 10 s is −247 J.

c)

The formula to calculate the work done is,

W 3 = F n scosθ

Here, the net force on the body is F n , the displacement of the body is s and the angle between the force and the displacement is θ.

Substitute the values in the above equation.

W 3 =( 5.04 )( 126 )cos0° ≈635 J

Thus, the work done by the net force in 10 s is 635 J.

d)

Since, the change in kinetic energy is equal to the work done. So

ΔKE= W 3

Substitute the values in the above equation.

ΔKE=635 J

Hence, the change in kinetic energy of the body in 10 s is 635 J.

According to the work energy theorem, the change in the kinetic energy is equal to the net work done by the body in the respective time.

Thus, the change in the kinetic energy is 635.04 J.