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Standard IX

Physics

Third Law of Motion

An object of ...

Question

An object of mass 10 kg moves at a constant speed of 10m/s. A constant force that acts for 4 sec on the object gives it a speed of 2m/s in opposite direction. The force acting on the object is:

-3N

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-30N

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30N

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Solution

The correct option is A -30N
Force is equal to the rate change of momentum. i.e $\to F = m \frac{d \to v}{d t}$
So, $F = 10 \times \frac{(- 2) - 10}{4} N = - 30 N$

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An object of mass 10 kg moves at a constant speed of 10m/s. A constant force that acts for 4 sec on the object gives it a speed of 2m/s in opposite direction. The force acting on the object is:

The correct option is A -30NForce is equal to the rate change of momentum. i.e →F=md→vdt So, F=10×(−2)−104N=−30N

The correct option is A -30N
Force is equal to the rate change of momentum. i.e $\to F = m \frac{d \to v}{d t}$
So, $F = 10 \times \frac{(- 2) - 10}{4} N = - 30 N$