Question

3) A body of mass 10 kg is pulled from rest with a constant force of 40 N. The force is applied for 2 sec. calculate (a) The distance travelled (b) Work done on the body (c) The KE of the body (d) The final velocity of the body​

Answer 1

Explanation:f=ma. S=ut+at(t)/2

s=at(t)/2 2s/t(t)=a

f=m(2s)/t(t)

f(t)(t)/2m=s

40(2)(2)/2(10)=s

160/20=s

s=8m

b.w=f(s)

40(8)=w. W=320J

KE=m(v)(v)/2

V(V)=U(u)+2aS

kE=M[(U)(u)+2as]

u=0

KE=2mas. KE=2(10)(4)(8). KE=640J

v(v)=u(u)+2as

v(v)=2(4)(8)

v(v)=64

v=8