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A bowling ball is dropped off a roof. The balls velocity when it hits is 6 m/a and it is brought to rest in .02 s. If the mass of the ball is 10kg, what force is needed to bring the ball to rest?

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Final answer:

To stop a 10kg bowling ball dropped from a roof with a velocity of 6 m/s in 0.02 s, a force of -3000 N is required. The negative sign represents the force's opposite direction to the ball's motion.

Step-by-step explanation:

To calculate the force needed to bring a 10kg bowling ball to rest when it hits the ground with a velocity of 6 m/s and is brought to rest in 0.02 seconds, we need to use the formula for force which is derived from Newton's second law of motion, that is, force = mass × acceleration. Here, we first find the acceleration by using the formula for acceleration (a = Δv/Δt), where Δv is the change in velocity and Δt is the change in time. Since the ball is moving at 6 m/s and is brought to rest, the change in velocity (Δv) is -6 m/s (since it's a deceleration, we take it as negative) and the change in time (Δt) is 0.02 s.

The acceleration is thus a = Δv/Δt = -6 m/s / 0.02 s = -300 m/s².

The magnitude of the force can now be calculated using F = ma, where m is the mass of the ball (10kg) and a is the acceleration just calculated:

F = 10kg × -300 m/s² = -3000 N.

The negative sign indicates the direction of the force is opposite to the direction of the velocity, which means it is an upward force that stops the ball.

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