Jonathan is riding a bicycle and encounters a hill of height 6.70 m. At the base of the hill, he is traveling at 5.00 m/s. When he reaches the top of the hill, he is traveling at 3.00 m/s. Jonathan and his bicycle together have a mass of 77.6 kg. Ignore friction in the bicycle mechanism and between the bicycle tires and the road.
(a) What is the total external work done on the system of Jonathan and the bicycle between the time he starts up the hill and the time he reaches the top?
I was thinking the answer is KE final minus KE initial, which is 620.8 Joules, but was not sure if this is correct or not.
(b) What is the change in potential energy stored in Jonathan's body during this process?
for part b, I thought this was Ug=mgh. I entered in -5095.2 Joules and that was incorrect. Why is it incorrect? On my second attempt, I changed the sign to positive and it is still incorrect. In fact, the problem says that it should be negative.
There is a part c to the question, but I got it correct. I have only one more attempt to the problem.
Help & an explanation would be great. If you only know how to do one part of the problem, please help me with that part of the problem that you do know.
Gravity does negative work on him
- m g h = -77.6* 9.81 * 6.7
= - 5100 Joules
That is how much potential energy he gains going up the hill.
He loses kinetic energy
loss in Ke = (1/2)(77.6)(25-9)
= 620 Joules
He needed to use 5100 Joules to get up the hill
620 of that came from the loss of kinetic energy as he slowed down (coasted)
The rest (5100-620) or 4480 Joules came from the power breakfast he ate earlier.
I'm still confused why the answer is -5100 Joules for part B when I got it wrong with -5095.2 Joules.
"The rest (5100-620) or 4480 Joules came from the power breakfast he ate earlier."
In other words he used 4480 Joules of energy that was stored in his body.
To answer part (a), we need to calculate the work done on the system of Jonathan and the bicycle. The work done is equal to the change in kinetic energy.
The initial kinetic energy (KE_initial) of the system is given by:
KE_initial = (1/2) * mass * velocity_initial^2
Substituting the given values:
KE_initial = (1/2) * 77.6 kg * (5.00 m/s)^2 = 970.0 J
The final kinetic energy (KE_final) of the system is given by:
KE_final = (1/2) * mass * velocity_final^2
Substituting the given values:
KE_final = (1/2) * 77.6 kg * (3.00 m/s)^2 = 346.8 J
The total work done on the system is then:
Work = KE_final - KE_initial
= 346.8 J - 970.0 J
= -623.2 J
So, the total external work done on the system of Jonathan and the bicycle is -623.2 Joules. Note that the negative sign indicates that work is done on the system.