# Work-Energy Help

Work-Energy!!Help

## Homework Statement

An object of mass 0.550kg is lifted from the floor to a height of 3.50m at a constant speed.
A) how much work is done by the lifting force?
W=fdotD w=mg*d=.550*9.8/3.5=18.87J is this right

B) how much work is done by the earth on the object?
this would be the same as a?

c)what is the net work done on the object?
W_net=W_applied-W_gravity=0?

D) What is the change in kinetic energy of the object?
v_f=v_i
W=0 so KE=0?

E) If the object is released from rest after it is lifted what is the kinetic energy just before it hits the floor? the velocity?
have no clue on this one
k_i+U_i=K_f+u_f
mgy=.5mv^2
v=8.28m/s
k=.5mv^2=18.87J?

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## Answers and Replies

All of your answers look good to me. For part e, you'll want to use the conservation of mechanical energy, since the only force acting on the object is the conservative gravitational force.

Edit: For part a it looks like you divided by 3.5m, it should be W = mg(d), no dividing necessary.

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