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Standard IX

Physics

Derivation of Velocity-Time Relation by Graphical Method

A constant re...

Question

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

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Solution

$F = m \times a$
$50 = 20 \times a$
$a = {2.5 m s}^{- 2}$
deacceleration is ${2.5 m s}^{- 2}$
$0 = 15 - 2.5 t$
$0 = 15 - 2.5 t$

$t = \frac{15}{2.5}$

$t = 6 s$

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Similar questions

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?

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

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 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

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

F=m×a 50=20×a a=2.5ms−2deacceleration is 2.5ms−20=15−2.5t0=15−2.5tt=152.5t=6s

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

$F = m \times a$
$50 = 20 \times a$
$a = {2.5 m s}^{- 2}$
deacceleration is ${2.5 m s}^{- 2}$
$0 = 15 - 2.5 t$
$0 = 15 - 2.5 t$

$t = \frac{15}{2.5}$

$t = 6 s$