Work and Energy

A 0.62 kg wood block is firmly attached to a very light horizontal spring (k=180 N/m). it is noted that the block-spring system, when compressed 5cm and released, stretches out 2.3cm beyond the equilibrium position before stopping and turning back. what is the coefficient of kinetic friction between the block and the floor?

Note by Jon-jon Castro
4 years, 11 months ago

No vote yet
1 vote

  Easy Math Editor

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold

- bulleted
- list

  • bulleted
  • list

1. numbered
2. list

  1. numbered
  2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1

paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
    # 4 spaces, and now they show
    # up as a code block.

    print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.
2 \times 3 \( 2 \times 3 \)
2^{34} \( 2^{34} \)
a_{i-1} \( a_{i-1} \)
\frac{2}{3} \( \frac{2}{3} \)
\sqrt{2} \( \sqrt{2} \)
\sum_{i=1}^3 \( \sum_{i=1}^3 \)
\sin \theta \( \sin \theta \)
\boxed{123} \( \boxed{123} \)

Comments

Sort by:

Top Newest

the potential energy stored in the spring block system at the initial stage will be 1/2kx^2 and this energy will be dissipated against the friction when its being released from the compressed configuration By energy conservation we get 1/2kx^2=U(kinetic) [since its not static} mg*(x+2.3) where x=5cm. solving we get u(kinetic)= 0.5073

Ramesh Goenka - 4 years, 11 months ago

Log in to reply

thank you so much..

Jon-jon Castro - 4 years, 11 months ago

Log in to reply

kindly answer please.. need some expert.. thanks a lot..

Jon-jon Castro - 4 years, 11 months ago

Log in to reply

×

Problem Loading...

Note Loading...

Set Loading...