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Q69PE

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College Physics (Urone)
Found in: Page 264

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Short Answer

A 105-kg basketball player crouches down 0.400 m while waiting to jump. After exerting a force on the floor through this 0.400 m, his feet leave the floor and his center of gravity rises 0.950 m above its normal standing erect position.

(a) Using energy considerations, calculate his velocity when he leaves the floor.

(b) What average force did he exert on the floor? (Do not neglect the force to support his weight as well as that to accelerate him.)

(c) What was his power output during the acceleration phase?

(a.) The velocity when the player leaves the floor is 4.32 m/s .

(b) The average force exerted by the player is 3472.88 N.

(c) The average power output during the acceleration is 7501.42 W.

See the step by step solution

Step by Step Solution

Step 1: Conservation of energy

The total energy of an isolated system always remains constant, only the form of energy can be changed.

When a body is thrown upwards, the kinetic energy of the body gets converted into potential energy of the body. Mathematically,

KE=PE

Step 2: Velocity when the player leaves the floor

(a)

When the player jumps, its total kinetic energy gets converted into kinetic energy i.e.,

12mvi2=mghf

Here, m is the mass of the player m=105 kg, vi is the initial velocity or the velocity when the player leaves the floor, g is the acceleration due to gravity g=9.8 m/s2 , and hf is height the center of gravity rises hf=0.950 m .

The expression for the velocity when the player leaves the floor is,

vi=2ghf

Putting all known values,

vi=2×9.8 m/s2×0.950 m=4.32 m/s

Therefore, the required velocity when the player leaves the floor is 4.32 m/s.

Step 3: Average force player exerts on the ground

(b)

The work done by the player during the jump is,

W=mghf-hiFd=mghf-hi

Here, F is the average force exerted by the player, d is the distance d=0.400 m , m is the mass of the playerm=105 kg , g is the acceleration due to gravity g=9.8 m/s2, hf is height the center of gravity rises hf=0.950 m , and hi is the initial height or distance the player crouches down hi=-0.400 m .

The expression for the average force is,

F=mghf-hid

Putting all known values,

F=105 kg×9.8 m/s2×0.950 m--0.400 m0.400 m=3472.88 N

Therefore, the required average force exerted by the player is 3472.88 N .

Step 4: Average power

(c)

The average power is,

P=Fvavg=F×vf+vi2

Here, F is the average force exerted by the player F=3472.88 N , vi is the initial velocity of the jump vi=4.32 m/s, and vf is the final velocity ( vf=0 at maximum height all the kinetic energy get converted into potential energy).

Putting all known values,

P=3472.88 N×0+4.32 m/s2=7501.42 W

Therefore, the required average power output during the acceleration is 7501.42 W .

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