Finding Velocity with Work-Energy Theorum

[Exuro89]

Homework Statement

There is an 8 kg block on a table with a 6 kg block hanging off of a pulley on the end of a table. The rope and pulley have negligible mass, and the pulley is frictionless. The coefficient of kinetic friction between the 8.00-kg block and the tabletop is 0.250 The blocks are released from rest.

Use energy methods to calculate the speed of the 6.00-kg block after it has descended 1.50 m.

Homework Equations

f=ma
KE=1/2mv^2

The Attempt at a Solution

My problem with this is all of the solutions I've seen use gravitational potential energy, however we haven't gotten to that chapter yet so I can't use it. How else would I solve for it? I know that the net force on the block is 58.8N from finding the smaller blocks force and subtracting the friction force from that.












Old problem

Homework Statement

A physics professor is pushed up a ramp inclined upward at θ=30.0º above the horizontal as he sits in his desk chair that slides on frictionless rollers. The combined mass of the professor and chair is m=85 kg. He is pushed S=2.50 m along the incline by a group of students who together exert a constant horizontal force of F=600 N. The professor’s speed at the bottom of the ramp is = 2.00 m/s. Using energy considerations, find the speed at the top of the ramp.

Homework Equations

W=K_2-K_1
F=ma
W=F*s

The Attempt at a Solution

First thing I did was find the net force on the box. The force of gravity along the displacement would be -mgsin(30) which is -416.5N. Combined with the positive 600N of constant force pushed on the chair we get a net force of 183.5N

Then I multiplied the force by the displacement to get 458.75J

Now I use the work-energy theorum to solve for velocity.

1/2mv_2^2 = 1/2mv_1^2 + Work_tot

1/2*85*v_2^2 = 1/2*85*2^2 + 458.75

v = 3.85m/s

I looked online to see if I could compare my answers with someone else and I found this site that has the same question, but it has a different answer and I'm not sure why mine doesn't match. http://www.hostos.cuny.edu/departments/math/MTRJ/archives/volume4/problemsolving/2.%20presentation.problem-solving.article.yr.pdf

Is there something I'm overlooking in the problem?

 

[PhanthomJay]

The students push with a horizontal force of 600 N. You assumed the force was parallel to the plane.

 

[Exuro89]

Ah I see, thanks. I have another question I'm having issues with

There is an 8 kg block on a table with a 6 kg block hanging off of a pulley on the end of a table. The rope and pulley have negligible mass, and the pulley is frictionless. The coefficient of kinetic friction between the 8.00-kg block and the tabletop is 0.250 The blocks are released from rest.

Use energy methods to calculate the speed of the 6.00-kg block after it has descended 1.50 m.



My problem with this is all of the solutions I've seen use gravitational potential energy, however we haven't gotten to that chapter yet so I can't use it. How else would I solve for it? I know that the net force on the block is 58.8N from finding the smaller blocks force and subtracting the friction force from that.

 

[PhanthomJay]

Find the work done by the net force and use the work-energy method. It is better to look at the work done by each force and add them, rather than find the net force and then the net work, since the term 'net force' is a bit unclear when each force acts in directions not in the same plane.

 

[Exuro89]

Okay, so the work friction does is u_k*m*g*s which would be -29.4J as the force is going the opposite of the displacement. Then the second block does mg*s which is 88.2J of work in the positive direction. Together that gives me a network of 58.8J in the positive direction. Now I can equal this to KE_2 - KE_1 and because its from rest it's just KE_2 which is 1/2mv^2 and solve for v? m in this equation needs to be both 6 and 8 correct, as this makes both blocks act as one?

I get v = 2.898

 

[PhanthomJay]

Okay, so the work friction does is u_k*m*g*s which would be -29.4J as the force is going the opposite of the displacement. Then the second block does mg*s which is 88.2J of work in the positive direction. Together that gives me a network of 58.8J in the positive direction. Now I can equal this to KE_2 - KE_1 and because its from rest it's just KE_2 which is 1/2mv^2 and solve for v? m in this equation needs to be both 6 and 8 correct, as this makes both blocks act as one?

I get v = 2.898

Yes, well done, round off to 2.9 m/s