The figure shows the velocity - time graph of a ball of mass $20 g$ moving along a straight line on a table. How much average force does the table exert on the ball to bring it to rest?

0.02 N

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0.03 N

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0.04 N

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0.05 N

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Solution

The correct option is C $0.04 N$
Average force,

$F = \frac{Δ p}{Δ t} = \frac{p_{f} - p_{i}}{Δ t} = \frac{m (v_{f} - v_{i})}{Δ t}$

$\Rightarrow F = \frac{20 \times 10^{- 3} (0 - 20)}{10} = - 0.04 N$

Negative sign shows that friction force applied by the table acts opposite to the initial velocity.

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The figure shows the velocity - time graph of a ball of mass 20 g moving along a straight line on a table. How much average force does the table exert on the ball to bring it to rest?

The correct option is C 0.04 NAverage force, F=ΔpΔt=pf−piΔt=m(vf−vi)Δt ⇒F=20×10−3(0−20)10=−0.04 N Negative sign shows that friction force applied by the table acts opposite to the initial velocity.

The figure shows the velocity - time graph of a ball of mass $20 g$ moving along a straight line on a table. How much average force does the table exert on the ball to bring it to rest?