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ICSE Class 9 Answered

A body of mass 500 g, initially at rest, is acted upon by a force which causes it to move a distance of 4 m in 2 s. Calculate the force applied. Sir i did it in two ways and got two answers. Ist way: m=500 g = 0.5 kg u=0 s=4m t=2s v=s/t = 4/2= 2 m/s From the 1st eqn. of motion, v = u + at a=(v - u)/t = 2-0/2= 1 m/s^2 F=ma   = 0.5×1 = 0.5 N 2nd way: m=500 g = 0.5 kg u=0 s=4m t=2s F=ma a=? From the 2nd eqn. of motion, s=ut + 1/2 at^2 4= 0 + 1/2 × a × 2×2 4=2a a=2m/s^2 F=ma    = 0.5 × 2 = 1 N Please tell me which way is correct. Kindly give me the explanations for the same.
Asked by kandappan | 29 Dec, 2019, 08:16: PM
answered-by-expert Expert Answer
Second method is correct way of solving this problem.
 
In First method, you used the relation v = s/t , which is meant for uniform motion,
 
But here force is acting on a mass that produces aceleration. Hence it is accelerated motion.
 
you can not use the relation , v = s/ t
 
whereas in second method, you did correctly using the relation s = ut +(1/2)at2
and calculated acceleration  and hence force.
 
So second method is correct.
Answered by Thiyagarajan K | 29 Dec, 2019, 10:32: PM
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