Answer to Question #118732 in Mechanics | Relativity for Navneel

Question

John is riding a bicycle and comes across a hill of height 7.30 m. At the base of the hill, he is traveling at 6.00 m/s. When he reaches the top of the hill, he is traveling at 1.00 m/s. Jonathan and his bicycle together have a mass of 85.0 kg. Ignore friction in the bicycle mechanism and between the bicycle tires and the road.
i) 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?
ii) What is the change in potential energy stored in Jonathan’s body during this process?

Answer 1

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 can be calculated from energy conservation principle.

At the base of the hill the system has initial kinetic and zero potential energy. The boy had to perform work to reach the top of the hill:

$$E_i=E_{K1}+W.$$

On top of the hill, he has potential energy and kinetic energy:

$$E_f=E_{P2}+E_{K2}.$$

Energy conservation principle states:

$$E_f=E_i,\\ \frac{mv_1^2}{2}+W=mgh+\frac{mv_2^2}{2},\\ \space\\ W=m(gh+0.5(v_2^2-v_1^2))=\\ =85(9.8\cdot7.3+0.5(1^2-6^2))=4593.4\text{ J}.$$

This is the work done by the system. The boy had to perform 4593 J of work to get to the top. The work done on the system is negative work done by the system: -4593.4 J.

The change in potential energy is the difference between the final and initial potential energies:

$$\Delta E_P=mgh_f-mgh_i=mg(h_f-h_i)=\\ =85\cdot9.8(7.3-0)=6080.9\text{ J}.$$