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A ball of play-doh with a mass of 1 kg is thrown at 5 m/s against a wall. The wall causes the play-doh to smash and come to a stop in 0.13 s. What force does the wall apply to the play-doh to make it stop?"

Provide the magnitude of the force applied by the wall (number only).

Respuesta :

xero099

The magnitude of the force applied by the wall on the ball while smashed is 38.462 newtons.

How to determine the required impact force to stop a ball of Play-Doh(TM)

After a quick reading of the statement we conclude that the impact theorem shall be useful to determine the dynamic force required to stop the ball when smashed. The impulse theorem results from the combination of the Newton's laws of motion, principle of energy conservation and work-energy theorem:

[tex]m \cdot v_{o} = F\cdot \Delta t[/tex]     (1)

Where:

  • m - Mass of the ball, in kilograms.
  • [tex]v_{o}[/tex] - Initial speed of the ball, in meters per second.
  • Δt - Impact time, in seconds.
  • F - Impact force, in newtons.

If we know that m = 1 kg, [tex]v_{o} = 5\,\frac{m}{s}[/tex] and Δt = 0.13 s, then the impact force is:

F = (1 kg) · (5 m/s)/(0.13 s)

F = 38.462 N

The magnitude of the force applied by the wall on the ball while smashed is 38.462 newtons. [tex]\blacksquare[/tex]

To learn more on impulse theorem, we kindly invite to check this verified question: https://brainly.com/question/11674698

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