Question

You lower a 5.10-kg box to the floor from a shelf 1.21m above the floor.
If you lower the box with a constant speed, what is the force you exert on the box? Is it upward or downward? (You must answer both parts to get credit.)
(in N)

A: 26.67

B: 31.21

C: 36.51

D: 42.72

E: 49.98

upward
downward

What is the work you do on the box?(make sure you get the right sign)
(in J)

A: -48.38

B: -60.48

C: -75.59

D: -94.49

E: -118.12

What is the magnitude of the change in potential energy of the box? Does it increase or decrease? (You must answer both correctly to get credit.)
(in J)

A: 41.91

B: 47.36

C: 53.52

D: 60.48

E: 68.34

Increase
Decrease

Consider what your answers would be if you lift the box back to the shelf. You do not need to enter the answers for this, but you do need to know what signs might change for the opposite case.

Answer 1

To calculate force required at constant speed, consider Free body diagram of box above:

For constant velocity, acceleration is zeo, hence force has to be balanced.
So,

F = mg

F =

F = 49.98 N upwards  

b) Work = where s is displacement moved by box and is angle between force and displacement.

Work =

Work = -60.48 J

c) Change in magnitude of potential energy = where h is change in height

=

=

As height decreases, so potential energy decreases.

Now if we lift the box.

a) Force will be in upwards direction and of same magnitude for zero acceleration as this force's will just be able to lift the box at constant velocity. This condition will be same as for lowering the box.

b) Work will be positive with same magnitude as while lifting force and displacement of box are in same direction, that is

cos 0 =1 . However, magnitude will be same as above as shelf's height is same as height by which it was lowered and same amount of force is being applied.

c) The potential energy will now increase by same magnitude as by which it decreased while lowering. (Mgh). Hence, height will increase here.