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Byju's Answer

Standard IX

Physics

The Equations of Motion

A body of mas...

Question

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?

3 second

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8 second

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6 second

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11 second

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Solution

The correct option is C
6 second

a = -50N/20kg = - 2.5m/s²

using first equation of motion

v-u = at

0-15 = -2.5 x t

t= 6 second

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Similar questions

A constant retarding force of 50 N is applied to a body of mass 20 kg moving initially with a speed of 15 ms^–1. How long does the body take to stop?

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 ?

Q. A constant retarding force of

50

N

is applied to a body of mass

20

k g

moving initially with a speed of

15

m s^{- 1}

. How long does the body take to stop?

Q. A constant retarding force of 50N is applied to a body of mass 10 kg moving initially with a speed of

10 m s^{- 1}

. The body comes to rest after

A .

A constant retarding force of 50 N applied to a body of mass 20 Kg moving initially with a speed of 15 m

s^{- 1}

. How long does the body take to stop?

B .

A constant force acting on a body of mass 3.0 Kg change its speed from 2.0 m

s^{- 1}

to 3.5 m

s^{- 1}

in 25 s. The direction of the motion of the body remain same, the magnitude of the force will be

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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?

The correct option is C 6 seconda = -50N/20kg = - 2.5m/s2 using first equation of motion v-u = at 0-15 = -2.5 x t t= 6 second

The correct option is C
6 second

a = -50N/20kg = - 2.5m/s²

using first equation of motion

v-u = at

0-15 = -2.5 x t

t= 6 second