Question

a) how much work is done in raising a 50.0 kg crate a distance of 1.5 m above a storeroom floor?
b) what is the change of potential energy as a result of this move?
c) how much kinetic energy will the crate have as it falls and hits the floor?

please show work

Answer 1

a) how much work is done in raising a 50.0 kg crate a distance of 1.5 m above a storeroom floor?

Work = Fd = mg h = 50.0 x 9.81x 1.5= 736 j (or 740 J using 2 S.F.)

b) what is the change of potential energy as a result of this move?

ΔPe = mg Δh= 50.0 x 9.81x 1.5= 740 J
c) how much kinetic energy will the crate have as it falls and hits the floor?

Kinetic energy at the very bottom is equal to potential energy at the very top

Pe_max= Ke_max = 740 J

please let me know if you have any questions.

Answer 2

(a)

 (b)

 (c) Since the PE lost is equal to the KE gained, then KE = 740 J.