Work-Energy Help: Solve for Object Lifted from Floor to Height

[MEAHH]

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?

 

[hotcommodity]

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.

 

[Moetasim]

I would like to clarify that work and energy are two different concepts. Work is the transfer of energy that occurs when a force is applied to an object and it causes displacement. Energy, on the other hand, is the ability to do work.

Now, let's address the questions one by one:

A) To calculate the work done by the lifting force, we use the formula W = Fd, where F is the lifting force and d is the distance moved. In this case, the lifting force is equal to the weight of the object (mg) and the distance moved is the height (3.50m). So, the work done by the lifting force is W = (0.550kg)(9.8m/s^2)(3.50m) = 18.87J. Your answer is correct.

B) The work done by the Earth on the object is equal to the negative of the work done by the lifting force. This is because the Earth is exerting a force on the object in the opposite direction of the lifting force. So, the work done by the Earth is -18.87J.

C) The net work done on the object is the sum of the work done by all forces acting on the object. In this case, there are only two forces: the lifting force and the force of gravity. So, the net work is W_net = W_applied + W_gravity = 18.87J + (-18.87J) = 0. This means that the total energy of the object remains the same.

D) The change in kinetic energy of the object is equal to the net work done on the object. Since we already calculated the net work to be 0, the change in kinetic energy is also 0. This means that the kinetic energy of the object remains the same.

E) If the object is released from rest after it is lifted, its kinetic energy just before it hits the floor will be equal to its potential energy at the top of its trajectory. This is because all of its potential energy is converted to kinetic energy as it falls. So, the kinetic energy just before it hits the floor will be equal to the potential energy at the top, which is mgh = (0.550kg)(9.8m/s^2)(3.50m) = 18.87J. The velocity just before it hits the floor can be calculated using the