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

1. Work done by the applied force in 10 s
2. Work done by friction in 10 s
3. Work done by the net force on the body in 10 s
4. Change in kinetic energy of the body in 10 s and interpret your results.

Answer 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 imparted to the body by the applied force is derived from Newton's second law of motion as follows:

`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 produced 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 the body covers is determined by the motion equation:

`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) Using the first equation of motion, the final velocity is 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`