Question

Problem 1: For all of the following problems, imagine a crane that is lifting a weight of mass m through a height h at constant speed, and consider the following three choices of system:

(1) system is the weight only; (2) system is the weight and the earth; (3) system is the crane only.

For each of these choices, and in this order, what is the net work done on the system by the external forces (Wext,sys) in the process considered?

(a) (1) 0; (2) mgh; (3) −mgh

(b) (1) mgh; (2) 0; (3) −mgh

(c) (1) −mgh; (2) mgh; (3) 0

(d) (1) 0; (2) mgh; (3) 0

Problem 2: For system (1) (weight only), which of the following is a correct statement of the energy changes going on inside the system during the process described?

(a) ∆Esource + ∆Ediss = −mgh

(b) ∆U G = mgh

(c) ∆K = 0

(d) ∆Ediss = mgh

Problem 3: For system (2) (weight and earth), which of the following is a correct statement of the energy changes going on inside the system during the process described?

(a) ∆Esource + ∆Ediss = −mgh

(b) ∆U G = mgh

(c) ∆K = 0

(d) ∆Ediss = mgh

Problem 4: For system (3) (crane only), which of the following is a correct statement of the energy changes going on inside the system during the process described?

(a) ∆Esource + ∆Ediss = −mgh

(b) ∆U G = mgh

(c) ∆K = 0

(d) ∆Ediss = mgh

Problem 5: If it takes a time ∆t to raise the weight through the height h, what is the power output of the crane for that process?

(a) mg (b) mgh (c) mgh/∆t (d) 0

Answer 1

1) if we consider system as weight only without considering earth, it implies there wont be any potential energy gain. Also since weight is moving at constant speed, there is no KE gain. => case 1) Wext ,sys = 0.

Now consider weight along with earth, it implies potential energy gain as mgh. =>case 2) Wext ,sys = mgh

Now consider crane system, it has lost energy = potntial energy of weight+earth system =-mgh => case 3) Wext ,sys = -mgh

2) for system (1) there is no PE gain and no KE gain as explained in Q.1 also there is no energy dissipated => correct answer option will be (c) ∆K = 0

3) for system (2) there is PE gain=mgh and no KE gain, aslso no energy is dissipated => correct answer option will be (b) ∆U G = mgh

4) for system (2) there is no PE gain and no KE gain, but energy dissipated = mgh => correct answer option will be (d) ∆Ediss = mgh

5) we know power = ∆E/∆t = mgh/∆t