4) Potential energy at the top of the building: 784 J
5) Potential energy halfway through the fall: 392 J
7) Kinetic energy just before hitting the ground: 784 J
Explanation:
4)
The potential energy of an object is the energy possessed by the object due to its position in the gravitational field; it is given by
[tex]PE=mgh[/tex]
where
m is the mass of the object
g is the acceleration due to gravity
h is the height of the object relative to the ground
For the bowling ball, when it sits on top of the building, we have:
m = 2 kg
[tex]g=9.8 m/s^2[/tex]
h = 40 m
Therefore, its potential energy is
[tex]PE=(2)(9.8)(40)=784 J[/tex]
5)
The potential energy of the ball when it is half way through the fall is:
[tex]PE=mgh'[/tex]
where
m = 2 kg is the mass of the ball
[tex]g=9.8 m/s^2[/tex] is the acceleration due to gravity
h' = 20 m is the height of the ball relative to the ground
Substituting into the equation, we find:
[tex]PE=(2)(9.8)(20)=392 J[/tex]
7)
The kinetic energy of the ball just before hitting the ground can be found by applying the law of conservation of energy.
In fact, the total mechanical energy of the ball during the fall is constant, and it is given by:
[tex]E=PE+KE[/tex]
where KE is the kinetic energy.
When the ball sits at the top of the building, KE = 0 (since the ball is not moving), so
[tex]E=PE=784 J[/tex]
When the ball is just about to hit the ground, its height is zero:
h = 0
This means that the potential energy is now zero:
PE = 0
And therefore, it means that all the mechanical energy is now kinetic energy, and so:
[tex]KE=E=784 J[/tex]
The kinetic energy can also be found by using the formula:
[tex]KE=\frac{1}{2}mv^2[/tex]
where
m = 2 kg is the mass of the ball
v = 28 m/s is the speed of the ball just before hitting the ground
Substituting,
[tex]KE=\frac{1}{2}(2)(28)^2=784 J[/tex]
Learn more about kinetic and potential energy:
brainly.com/question/6536722
brainly.com/question/1198647
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