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1) A 2.5 kg ball strikes a wall with a velocity of 8.5 m/s to the left. The ball bounces off with a
velocity of 7.5 m/s to the right. If the ball is in contact with the wall for 0.25 s, what is the
constant force exerted on the ball by the wall?

Respuesta :

Lanuel

Answer:

Force = -10 Newton.

Explanation:

Given the following data;

Initial velocity = 8.5m/s

Final velocity = 7.5m/s

Time = 0.25 seconds

Mass = 2.5kg

To find the force exerted;

First of all, we would determine the acceleration of the ball;

Acceleration = (v - u)/t

Acceleration =(7.5 - 8.5)/0.25

Acceleration = -1/0.25

Acceleration = -4 m/s²

Now, to find the force exerted;

Force = mass * acceleration

Force = 2.5 * -4

Force = -10 Newton

Note: The force is negative because it is an opposite reaction.

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