Q. 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

Solution:

Using $S_n = u +\frac{a}{2}(2n - 1), $ we get $5 = u + \frac{a}{2}(2 \times 1 - 1)$ i.e. 10 = 2u + a and
$2 = u + \frac{a}{2} (2 \times 3 - 1) $ i.e. 4 = 2u +5a
Using these equations, we have
4a = - 6 i.e. $a = - \frac{6}{4} = - \frac{3}{2} \, m \, s^{-2}$
Then. force. |F| = ma = 4 $\times \frac{3}{2}$ = 6 N

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