A body of mass 4 ~kg is accelerated upon by a constant force, travels a distance of 5 ~m in the 1st second and a distance of 2 ~m in the 3 rd secon(d)The force acting on the body is

Step-by-step solution

We know that, displacement travelled by a moving body with constant acceleration a in n th second, s_ n =u+ 1 2 a(2 n-1) For first second, n=1 aligned s₁ &=u+ 1 2 a(2 1-1) 5 &=u+ a 2 ( i ) [ s₁=5 ~m ] aligned For third second, n=3 array rlr s₃ & =u+ a 2 (2 3-1) 2 & =u+ 5 a 2 & ( ii) [ s₃=2 ~m ] . array Subtracting Eq. (i) from Eq. (ii), we get aligned &-3=2 a & a= -3 2 ~m / s ² aligned Force acting on the body, F=m a=4 ( -3 2 )=-6 ~N Hence, force acting on the body in opposite direction of its motion will be 6 ~N .