Question

A force of 5.0 N acts on a 15 kg body initially at rest. Compute the work done by the force in (a) the first, (b) the second, and (c) the third seconds and (d) the instantaneous power due to the force at the end of the third second.

Answer 1

Answer:

(a) 0.833 j

(b) 2.497 j

(c)  4.1625 j

(d) 4.995 watt      

Explanation:

We have given force F = 5 N

Mass of the body m = 15 kg

So acceleration $a=\frac{F}{m}=\frac{5}{15}=0.333m/sec^2$

As the body starts from rest so initial velocity u = 0 m/sec

(a) From second equation of motion $s=ut+\frac{1}{2}at^2$

For t = 1 sec

$s=0\times 1+\frac{1}{2}\times 0.333\times 1^2=0.1666m$

We know that work done W =force × distance = 5×0.1666 =0.833 j

(b) For t = 2 sec

$s=0\times 2+\frac{1}{2}\times 0.333\times 2^2=0.666m$

We know that work done W =force × distance = 5×0.666 =3.33 j

So work done in second second = 3.33-0.833 = 2.497 j

(c) For t = 3 sec

$s=0\times 3+\frac{1}{2}\times 0.333\times 3^2=1.4985m$

We know that work done W =force × distance = 5×1.4985 =7.4925 j

So work done in third second = 7.4925 - 2.497 -0.833 = 4.1625 j

(d) Velocity at the end of third second v = u+at

So v = 0+0.333×3 = 0.999 m /sec

We know that power P = force × velocity

So power = 5× 0.999 = 4.995 watt