The Work - Energy Theorem in a pulley?

  • #1
Consider the system shown in the figure. 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 [tex]\mu_k=0.250.[/tex] 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.

[tex]\Delta * K[/tex] = [tex](1/2)(m)(v_f^2 - v_i^2)[/tex]

But, it was at rest, so [tex](1/2)(m)(v_f^2)[/tex] is what matters.

What do I do next?
 

Attachments

  • Imajj.jpg
    Imajj.jpg
    9.8 KB · Views: 528

Answers and Replies

  • #2
Fermat
Homework Helper
872
1
You will have a loss in PE balanced by a gain in KE and work done.

What is it that loses PE, and by how much ?

What are the gains in KE and by how much ?

What work is done and how much ?
 
Last edited:

Related Threads on The Work - Energy Theorem in a pulley?

Replies
9
Views
9K
  • Last Post
Replies
3
Views
14K
  • Last Post
Replies
5
Views
5K
  • Last Post
Replies
2
Views
3K
  • Last Post
2
Replies
27
Views
2K
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
8
Views
1K
  • Last Post
Replies
4
Views
1K
  • Last Post
Replies
6
Views
4K
Top