Request a call back

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
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

## Application Videos

ICSE 9 - Physics
Asked by modhukuruvenkateshwarlu | 13 Apr, 2024, 03:03: PM
ICSE 9 - Physics
Asked by thesangeeta1990 | 08 Mar, 2024, 04:37: PM
ICSE 9 - Physics
Asked by navyaindu48 | 21 Sep, 2023, 07:57: PM
ICSE 9 - Physics
Asked by krashish57 | 14 Sep, 2023, 11:05: AM
ICSE 9 - Physics
Asked by rishil23dec | 08 Aug, 2023, 09:31: PM
ICSE 9 - Physics
Asked by msrikanta11 | 13 Jul, 2023, 09:42: AM