Question

A body of mass m is resting on a horizontal rough surface.A force of F 20N is applied to it for 10s parallel to the surface. Find K.E after 10s.Coefficient of kinetic friction is 0.3.

Answer 1

Explanation:

F=20N

N(K)=0.3

F=N(K)mg

20=0.3×m×9.8

m=20/0.3×9.8=6.802Kg

t=10s

F=ma

20=6.802×a

a=2.94m/s²

t=10s,u=0

then

a=v-u/t

2.94=v-0/10

v=29.4m/s

K.E=mv²/2

K.E=6.082×29.4×29.4/2=2628.51JOULES