Question

6.2 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​

Answer 1

Answer:

t=10s

u=0m/s

F=7N

m=2kg

a= 7/2=3.5m/s²

frictional force,f= 0.1xg=-1N(g=10m/s²)

a'=1/2=-0.5m/s²(acceleration due to friction)

so total acceleration,A=3.5-0.5=3m/s²

S=ut+1/2At²

= 1/2x3x100=150m

Work done=7x150=1050 J