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nekit [7.7K]
3 years ago
5

A 2.00-kilogram mud ball drops from rest at a height of 17.0 m. If the impact between the ball and the ground lasts 0.46 s, what

is the magnitude of the average force exerted by the ball on the ground
Physics
1 answer:
Softa [21]3 years ago
5 0

Answer: 79.35 N

Explanation: according to the impulse momentum theorem,

Impulse = change in momentum

Where impulse = force × time and change in momentum = m ( v - u).

The object was initially at rest, hence it initial velocity is zero.

To get the final velocity, we use the formulae below

v² = u² + 2gh

Where h = height of the cliff = 17m

v² = 2 × 9.8 × 17

v² = 333.2

v = √333.2

v = 18.25 m/s

At t = 0.46s and v = 18.25 m/s, we can get the average force of impact

F×0.46 = 2 (18.25 - 0)

F × 0.46 = 2 (18.25)

F × 0.46 = 36.5

F = 36.5 /0.46 = 79.35 N

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A skateboarder is skating back and forth on the halfpipe as seen below. As he skates his energy transforms from potential energy
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Answer:

Friction and air resistance cause some of his kinetic energy to be “lost”. This makes him slow down.

Explanation:

The law of conservation of energy states that in absence of frictional forces, the mechanical energy of an object (given by the sum of its kinetic and potential energy) is conserved. In such a situation, the skateboarder would never stop his motion, because potential energy is continuously converted into kinetic energy and vice-versa, but the total energy remains the same so he would never stop.

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Friction and air resistance cause some of his kinetic energy to be “lost”. This makes him slow down.

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