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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

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4 s

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6 s

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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

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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