You visited us 2 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard IX

Physics

Second Law of Motion

A constant re...

Question

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

2 s

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

4 s

No worries! We‘ve got your back. Try BYJU‘S free classes today!

6 s

No worries! We‘ve got your back. Try BYJU‘S free classes today!

8 s

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A 2 s
Here, F= -50 N (-ve sign for retardation) m=10 kg, u= 10 $m s^{- 1}$ , v=0

As F=ma $∴ a = \frac{F}{m} = \frac{- 50 N}{10 k g} = - 5 m s^{- 2}$

Using v=u+at $∴ t = \frac{v - u}{a} = \frac{0 - 10 m s^{- 1}}{- 5 m s^{- 2}}$ =2s

Suggest Corrections

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?

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?

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

Q. A body of mass '

m

' is travelling with a velocity '

u

'. When a constant retarding force '

F

' is applied, it comes to rest after travelling a distance '

s_{1}

'. If the initial velocity is '

2 u

', with the same force '

F

', the distance travelled before it comes to rest is '

s_{2}

'. Then

Join BYJU'S Learning Program

A constant retarding force of 50N is applied to a body of mass 10 kg moving initially with a speed of 10ms−1. The body comes to rest after

The correct option is A 2 sHere, F= -50 N (-ve sign for retardation) m=10 kg, u= 10 ms−1, v=0As F=ma ∴a=Fm=−50N10kg=−5ms−2Using v=u+at ∴t=v−ua=0−10ms−1−5ms−2=2s

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

The correct option is A 2 s
Here, F= -50 N (-ve sign for retardation) m=10 kg, u= 10 $m s^{- 1}$ , v=0

As F=ma $∴ a = \frac{F}{m} = \frac{- 50 N}{10 k g} = - 5 m s^{- 2}$

Using v=u+at $∴ t = \frac{v - u}{a} = \frac{0 - 10 m s^{- 1}}{- 5 m s^{- 2}}$ =2s