show all work A 7 kg ball (initial vertical velocity = 0m/s) was dropped from a
ID: 1461935 • Letter: S
Question
show all work A 7 kg ball (initial vertical velocity = 0m/s) was dropped from a height of 12 m. a) What was the potential, kinetic, and total mechanical energy of the ball at the instant it was dropped (height = 12m, velocity = 0m/s)?(3pts) potential = kinetic = Total = b) What was the total mechanical energy of the ball 0.2 second after it was dropped? (Hint: Conservation of energy) (1pt) answer= c) What was the potential, kinetic, and total mechanical energy of the ball when the ball was dropped to a height of 1m? (Hint: Conservation of energy. You can calculate the kinetic energy of the ball based on the total mechanical energy and the potential energy) (3pts) Potential Answer= Kinetic = Total = d) You started to decelerate the ball (i.e. ball became in contact with your hands) when it was 1m above ground, and the ball came to a complete stop (velocity = 0) when it was 0.6m above ground. What was the potential, kinetic, and total mechanical energy of the ball when it came to a stop? (Height = 0.6m, velocity = 0m/s) (Hint: Once you touch the ball (apply force to the ball), the ball’s energy is no longer conserved) (3pt) potential = kinetic= total= e) How much work did you do on the ball to bring it to a stop? (Hint: work = change in total mechanical energy) (1pt) (Hint: how did the total mechanical energy change from c to d?) answer= f) What was the average force you exerted on the ball to bring it to a stop? (Hint: Work = force x displacement) (1pt) answer= g) In a few complete sentences, describe how the potential, kinetic, and total mechanical energy of the ball changed when it was free-falling from 10m to 1m. (3pts) h) In a few complete sentences, describe how the potential, kinetic, and total mechanical energy of the ball changed from the time you made contact with the ball to the time the ball came to a complete stop. (3pts) Answer
Explanation / Answer
Solution: Given Mass of the ball m = 7 kg
Initial vertical velocity v = 0 m/s
Height above the ground h = 12 m
a) What was the potential, kinetic, and total mechanical energy of the ball at the instant it was dropped (height = 12m, velocity = 0m/s)?(3pts) potential = kinetic = Total =
Solution: Gravitational potential energy of the ball
U = m*g*h
U = (7kg)*(9.81m/s2)*(12m)
U = 824.04 J
Kinetic energy of the ball
K = (1/2)*m*v2
K = (1/2)*(7kg)*(0m/s)2
K = 0 J
Total energy of the ball T is,
T = K+U -------------------------------------------------------------(1)
T = 0 J + 824.04J
T = 824.04 J
Thus the answers are Potential energy U = 824.04 J, Kinetic energy K = 0 J and total energy T = 824.04 J
-----------------------------------------------------------------------------------------------------------------------------------------
b) What was the total mechanical energy of the ball 0.2 second after it was dropped? (Hint: Conservation of energy) (1pt) answer=
Solution: Conservation of the total energy states that the sum of the kinetic energy and the potential energy do not change in the absence of dissipative forces. In other words total energy does not change, T = constant.
In our case, T = 824.04 J
Thus total mechanical energy of the ball 0.2 s after it was dropped is 824.04 J
-------------------------------------------------------------------------------------------------------------------------------------
c) What was the potential, kinetic, and total mechanical energy of the ball when the ball was dropped to a height of 1m? (Hint: Conservation of energy. You can calculate the kinetic energy of the ball based on the total mechanical energy and the potential energy) (3pts) Potential Answer= Kinetic = Total =
Solution: Total energy T = 824.04 J
When the ball is h’= 1 m above the ground, it potential energy is
U’ = mgh’
U’ = (7kg)*(9.81m/s2)*(1m)
U’ = 68.67 J
Thus kinetic energy K can be found from equation (1) in this case it is T = K’ + U’,
K’ = T – U’
K’ = 824.04 J – 68.67 J
K’ = 755.37 J
Total energy T do not change, thus
T = 824.04 J
Thus the answers are, potential energy = 68.67 J, Kinetic energy = 755.37 J and total energy = 824.04 J
-------------------------------------------------------------------------------------------------------------------------------------
d) You started to decelerate the ball (i.e. ball became in contact with your hands) when it was 1m above ground, and the ball came to a complete stop (velocity = 0) when it was 0.6m above ground. What was the potential, kinetic, and total mechanical energy of the ball when it came to a stop? (Height = 0.6m, velocity = 0m/s) (Hint: Once you touch the ball (apply force to the ball), the ball’s energy is no longer conserved) (3pt) potential = kinetic= total=
Solution: When the ball is stopped at the height of 0.6 m above the ground, its velocity v is zero, thus kinetic energy when it is stopped is,
K” = (1/2)*m*v2
K” = (1/2)*(7kg)*(0m/s)2
K” = 0 J
Potential energy of the ball h”= 0.6 m above the ground is,
U” = m*g*h”
U” = (7kg)*(9.81m/s2)*(0.6m)
U” = 41.202 J
The new total mechanical energy T” is
T” = K” + U”
T” = 0 J + 41.202 J
T” = 41.202 J
Thus the answers are, kinetic energy = 0 J, potential energy = 41.202 J and total energy = 41.202 J
----------------------------------------------------------------------------------------------------------------------------------
e) How much work did you do on the ball to bring it to a stop? (Hint: work = change in total mechanical energy) (1pt) (Hint: how did the total mechanical energy change from c to d?) answer=
Solution: Work done by the hand in stopping the ball = change in total mechanical energy
W = T – T”
W = 824.04 J – 41.202 J
W = 782.838 J
Thus the work of 782.838 J was done on the ball.
------------------------------------------------------------------------------------------------------------------------------------
f) What was the average force you exerted on the ball to bring it to a stop? (Hint: Work = force x displacement) (1pt) answer=
Solution: When the ball was at 1m above the ground, the contact with hand stops it in 0.4 m. Thus it is stopped at 0.6 m above the ground. Thus the displacement during the application of the force is x = 0.4 m
We have
W = Favg*x
Favg = W/x
Favg = (782.838 J)/(0.4m)
Favg = 1957.095 N
Thus the average force exerted by the hand on the ball is 1957.095 N.
------------------------------------------------------------------------------------------------------------------------------
g) In a few complete sentences, describe how the potential, kinetic, and total mechanical energy of the ball changed when it was free-falling from 10m to 1m. (3pts)
Solution: The total mechanical energy of the ball must be constant during the free fall. When the ball starts falling with zero velocity, it has maximum potential energy and zero kinetic energy. As the ball gains speed while it is falling, the decrement in the potential energy causes increment of the ball’s kinetic energy exact amount at any point between 10 m to 1 m. This is why total mechanical energy remains the same or do not change during the free fall of the ball.
------------------------------------------------------------------------------------------------------------------------------------------------------------------
h) In a few complete sentences, describe how the potential, kinetic, and total mechanical energy of the ball changed from the time you made contact with the ball to the time the ball came to a complete stop. (3pts) Answer
Solution: When the hand makes the contact with the falling ball at the height of 1m above the ground, it has kinetic energy K = 755.37 J and potential energy of U = 68.67 J.
During the displacement of 0.4 m, potential energy of mgh = (7kg)*(9.81m/s2)*(0.4m) = 27.468 J is converted into the kinetic energy of the ball. Thus we do work against (755.37 J + 27.468 J = 782.838 J which tallies with the answer of part part (e).
Thus since the contact till stopping, the kinetic energy becomes zero, potential energy decreses due to decrease in height and total energy also decreases as the kinetic energy of the ball become zero and potential energy id further redueced due to decreases in height.
After stopping, the final total energy of the ball is
Tfinal = Kfinal + Ufinal
Tfinal = 0 J + 41.202 J
Tfinal = 41.202 J.
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.