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The velocity versus time graph of a ball of mass $50 g$ rolling on a concrete floor is shown in the figure. The acceleration and frictional force of the floor on the ball respectively

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10 m / s^{2} and 5 N

- 10 m / s^{2} and 5 N

10 m / s^{2} and 0.5 N

- 10 m / s^{2} and 0.5 N

answer is D.

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Detailed Solution

The velocity versus time graph of a ball of mass

50 g

rolling on a concrete floor is shown in the figure. The acceleration and frictional force of the floor on the ball are given by

- 10 m / s^{2} and 0.5 N

As we know that Acceleration,

a = \frac{v - u}{t 2 - t 1}

According to the given graph, we have information as
Initial velocity

u = 80 m / s

Final Velocity

v = 0 m / s

Time for the initial

t_{1} = 0 s,

Time for the final

t_{2} = 8 s

Hence,

a = \frac{0 - 80}{8 - 0} = - 10

m/s2 Now, the frictional force

F = ma

\Rightarrow F = 50 \times 1 0^{- 3}

kg ×

10

m/s2

\Rightarrow F = 0.5

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