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Artist 52 [7]
3 years ago
7

A 0.537-kg basketball is dropped out of a window that is 5.88 m above the ground. The ball is caught by a person whose hands are

1.98 m above the ground. (a) How much work is done on the ball by its weight? What is the gravitational potential energy of the basketball, relative to the ground, when it is (b) released and (c) caught? (d) What is the change (PEf - PE0) in the ball's gravitational potential energy?
Physics
1 answer:
jeka57 [31]3 years ago
5 0

Answer:

a) 20.54 J

b) 30.97 J

c) 10.43 J

d) -20.54 J

Explanation:

m = Mass

g = Acceleration due to gravity = 9.81 m/s²

h = Height

W=mgh\\\Rightarrow W=0.537\times 9.81\times (5.88-1.98)\\\Rightarrow W=20.54\ J

Work done by the ball's weight is 20.54 J

W=mgh\\\Rightarrow W=0.537\times 9.81\times (5.88)\\\Rightarrow W=30.97\ J

Gravitational potential energy of the basketball, relative to the ground, when it is released is 30.97 J

W=mgh\\\Rightarrow W=0.537\times 9.81\times (1.98)\\\Rightarrow W=10.43\ J

Gravitational potential energy of the basketball, relative to the ground, when it is released is 10.43 J

Change in gravitational potential energy

\Delta U=10.43-30.97=-20.54\ J

Change in gravitational potential energy is given by -20.54 J

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What force must the deltoid muscle provide to keep the arm in this position?
ruslelena [56]

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Additional Information:

Some numerical information are missing from the question. However, I will derive the formula to calculate the force of the deltoid muscle. All you need to do is insert the necessary information and calculate.  

Explanation:

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2. The torque of the arm, T_{arm}  

Assuming the arm is just being stretched and there is no rotation going on,

                        T_{Deltoid} = 0

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                  r_{d}F_{d}sin\alpha_{d} = r_{a}F_{a}sin\alpha_{a}

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r_{d} is radius of the deltoid

F_{d} is the force of the deltiod

\alpha_{d} is the angle of the deltiod

r_{a} is the radius of the arm

F_{a} is the force of the arm , F_{a} = mg  which is the mass of the arm and acceleration due to gravity

\alpha_{a} is the angle of the arm

The force of the deltoid muscle is,

                                 F_{d} = \frac {r_{a}F_{a}sin\alpha_{a}}{r_{d}sin\alpha_{d}}

but F_{a} = mg ,

                ∴            F_{d} = \frac {r_{a}mgsin\alpha_{a}}{r_{d}sin\alpha_{d}}

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