#### 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.

#### Solution 1

Mass of the body, *m* = 2 kg

Applied force, *F* = 7 N

Coefficient of kinetic friction, *µ* = 0.1

Initial velocity,* u* = 0

Time, *t* = 10 s

The acceleration produced in the body by the applied force is given by Newton’s second law of motion as:

`a' =F/m = 7/2 = 3.5 "m/s"^2`

Frictional force is given as

`f = mumg`

`= 0.1 xx 2xx 9.8 = - 1.96 N`

The acceleration producd by the frictional force

a"`= 1.96/2 = -0.98 "m/s"^2`

Total acceleration of the body:

a = a' + a"

`=3.5 + (-0.098) = 2.52 "m/s"^2`

The distance travelled by the body is given by the equation of motion:

`s = ut + 1/2 at^2`

`= 0 + 1/2 xx 2.52 xx (10)^2

= 126 m`

**a)** Work done by the applied force, *W*_{a} = *F* ×* s* = 7 × 126 = 882 J

**(b)** Work done by the frictional force,* W*_{f}_{ =} *F* ×* s* = –1.96 × 126 = –247 J

**(c) **Net force = 7 + (–1.96) = 5.04 N

Work done by the net force, *W*_{net}= 5.04 ×126 = 635 J

**(d)** From the first equation of motion, final velocity can be calculated as:

*v* = *u* + *at*

= 0 + 2.52 × 10 = 25.2 m/s

Change in kinetic energy = `1/2 mv^2 - 1/2 mu^2`

`= 1/2xx2(v^2-u^2) = (25.2)^2- 0^2 = 635 J`

#### Solution 2

(a) We know that U_{k} = frictional force/normal reaction frictional force = U_{k} x normal reaction

= 0.1 x 2 kg wt = 0.1 x 2 x 9.8 N = 1.96 N

net effective force = (7 – 1.96) N = 5.04 N

acceleration = 5.04/2 ms^{-2} = 2.52 ms^{-2}

distance, s=1/2x 2.52 x 10 x 10 = 126 m

work done by applied force = 7 x 126 J = 882 J

(b) Work done by friction = 1.96 x 126 = -246.96 J

(c) Work done by net force = 5.04 x 126 = 635.04 J

(d) Change in the kinetic energy of the body = work done by the net force in 10 seconds = 635.04 J (This is in accordance with work-energy theorem).