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, 20:16: 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, 22:32: PM
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