Question

A body of mass 1 kg is moving with a velocity of

100 cms–1. It is brought to rest in 10 s by applying

a force against its motion. What distance it moved

through before coming to rest? What is the

magnitude of the retarding force?
PLEASE answer if you know step by step

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Answer 1

m =1 kg

u = 100 cms^-1

= 1 ms^-1

v=0

t =10 s

s = ?

Now, a =v-u /t

= 0-100/10

= -10 m/s^-2

Now, s = ut +1/2 at^2

= 0×10+1/2×-10 ×10^2

= 0 - 5×100

= -500 m

Now, Force(F) = ma

= 1×10

= 10 N

Answer 2

Answer:

s=5 m and F=0.1 N

Step-by-step explanation:

m= 1 kg

u=100cm/s. =1m/s

v=0 m/s

t=10 sec

We have

a= (v-u)/t. =(0-1)/10

=-0.1 m/s²

Now ,

v²=u²+2as

0=1-2*0.1*s

0.2s=1

s=5 m

Also,

regarding force (F)=ma

=1*0.1

=0.1 N