Momentum of an object changes from $5 k g m / s$ to $15 k g m / s$ in $2$ seconds. What is the force applied on the object?

10 N

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5 N

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20 N

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40 N

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Solution

The correct option is C $5 N$
$F = \frac{Δ p}{Δ t} = \frac{(p 2 - p 1)}{t} = \frac{(15 - 5)}{2} = 5 N$

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Momentum of an object changes from 5 kg m/s to 15 kg m/s in 2 seconds. What is the force applied on the object?

The correct option is C 5 NF=ΔpΔt=(p2−p1)t=(15−5)2=5 N

Momentum of an object changes from $5 k g m / s$ to $15 k g m / s$ in $2$ seconds. What is the force applied on the object?

The correct option is C $5 N$
$F = \frac{Δ p}{Δ t} = \frac{(p 2 - p 1)}{t} = \frac{(15 - 5)}{2} = 5 N$