Question

A body of mass 4 kg is accelerated upon by a constant force,
travels a distance of 5 m in the first second and a distance of
2 m in the third second. The force acting on the body is
[KCET 2008]
(a) 6N
(b) 8N
(c) 2N
(d) 4N​

Answer 1

Answer:

(a)6N

Explanation:

Given,m=4 kg,S1=5m,S2=2m

Distance traveled in nth second S=u+1/2a(2n-1)

S1=5=u+1/2a(2(1)-1)=u+1/2a_______1

S2=2=u+1/2a(2(3)-1)=u+5/2a———2

From 1 and 2,

We get a=-1.5m/s²,u=3.75m/s

F=ma=4*(-1.5)=-6kN