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
8N
6N
2N
4N

Answer 1

Dear Student,

As the force acting on the body is constant, its acceleration will also be constant.

Let initial velocity='u'

Let acceleration='a'

The distance travelled by a body undergoing consant acceleration in the n^th second is given by

S_n=u + (a/2)(2n-1)-----------(1)

Hence according to question, S₁=5m and S₃=2m

From (1),

S₁=u + (a/2)[2(1)-1]

or 5=u + a/2----------(2)

Also, S₃=u + (a/2)[2(3)-1]

or 2=u + (5/2)a-----------(3)

Solving (2) and (3) for 'u' and 'a' we get,

u=33/6 and a=(-3/2)

Hence acceleration=a=-3/2 ms^-2.

Mass of body = m = 4kg

Let force=F

Hence by 2nd law of motion,

F = ma

or F = 4(-3/2)

or F = -6N

Hence the magnitude of force is 6N and is acting opposite to the direction of motion of the body.

Regards,