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Question

A constant retarding force of 50 N is applied to a body of mass 20 kg moving initially with a speed of 15 ms1. How long does the body take to stop?

Solution
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Retarding force, F=50N
Mass of the body, m=20kg
Initial velocity of the body, u=15m/s
Final velocity of the body, v=0

Using Newtons second law of motion, the acceleration (a) produced is given by:
F=ma
50=20 ×a
a=50/20=2.5m/s2
Using the first equation of motion, the time (t) taken by the body to come at rest
v=u+at
t=u/a=15/2.5=6s

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Q1
A constant retarding force of 50 N is applied to a body of mass 20 kg moving initially with a speed of 15 ms1. How long does the body take to stop?
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A body of mass 20 kg is moving initially with a speed of 15 m/s. If a constant retarding force of 50 N is applied on it, how long does the body take to stop?


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A. A constant retarding force of 50 N applied to a body of mass 20 Kg moving initially with a speed of 15 m s1. How long does the body take to stop?
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Q5

A constant retarding force of 50N is applied to a body of mass 20kg moving with the speed of 15m/s. How long does the body take to stop?

Ans1 Given

F=50N

M=20kg

u=15m/s

v=0

As F=ma

a=50/20=2.5m/s^2

Since the retarding is negativd so -2.5 m/s^2

Time=

v=u+at

=0-15/-2.5=6s

Ans2 F=ma

50=20×0-15/t

50=20×-15/t

50/20×-15=1/t

-6 s=t

Why the 2 is incorrect ?

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