# A block collides with a horizontal spring

• ohhi
In summary, a 2.7 kg block collides with a horizontal spring of 356 N/m. The block compresses the spring by 13.0 cm and has a coefficient of kinetic friction of 0.24. The work done by the spring in bringing the block to rest can be calculated using the equations W = Fd and Us = 0.5kx^2, which results in a value of 3.0. It is important to take into account the force of friction when calculating the work done by the spring.
ohhi

## Homework Statement

A moving 2.7 kg block collides with a horizontal spring whose spring constant is 356 N/m.

The block compresses the spring a maximum distance of 13.0 cm from its rest postion. The coefficient of kinetic friction between the block and the horizontal surface is 0.24. What is the work done by the spring in bringing the block to rest?

F=kx
w=F*S
f_k=mu_k * n

## The Attempt at a Solution

F= 356 N * .13 m = 46.28 N, so the box exerts 46.28 N on the spring.

The force from friction is .24 * 2.7 kg * 9.80 m/s = 6.35.
From here I am unsure of where to go. I was under the impression that I'd simple multiply 46.28 N * .13 m, but that isn't giving me the correct answer. Do I need to also subtract the work done by friction? I had assumed that was taken into account by removing the 6.35. Input would be greatly appreciated.

Edit: Am I simply getting it wrong because I'm not putting in a negative sign?

Last edited:
Use energies ...
See the work related equations:
W = Fd
W = change in U
Us = 0.5kx^2 ...

rootX said:
Use energies ...
See the work related equations:
W = Fd
W = change in U
Us = 0.5kx^2 ...

So, Us= 3.0 = W?

## What is the relationship between a block and a horizontal spring?

The block and horizontal spring have a physical relationship in which the block collides with the spring and causes it to compress or stretch.

## What happens to the energy during a block's collision with a horizontal spring?

During the collision, the energy of the block is transferred to the spring in the form of potential energy. The spring then exerts a force on the block, causing it to move or compress.

## How does the mass of the block affect the motion during a collision with a horizontal spring?

The mass of the block affects the amplitude and frequency of the resulting motion. A heavier block will cause the spring to compress more and have a lower frequency compared to a lighter block.

## What factors determine the maximum compression or stretch of a horizontal spring during a collision?

The maximum compression or stretch of a horizontal spring during a collision is determined by the initial velocity and mass of the block, as well as the spring constant and any external forces acting on the system.

## Can a horizontal spring be used to calculate the initial velocity of a block?

Yes, the spring constant and the maximum compression or stretch of the spring can be used to calculate the initial velocity of the block using the formula v = √(kx²/m), where v is the initial velocity, k is the spring constant, x is the maximum compression or stretch, and m is the mass of the block.

• Introductory Physics Homework Help
Replies
13
Views
957
• Introductory Physics Homework Help
Replies
27
Views
6K
• Introductory Physics Homework Help
Replies
9
Views
2K
• Introductory Physics Homework Help
Replies
10
Views
1K
• Introductory Physics Homework Help
Replies
5
Views
1K
• Introductory Physics Homework Help
Replies
16
Views
2K
• Introductory Physics Homework Help
Replies
1
Views
2K
• Introductory Physics Homework Help
Replies
2
Views
888
• Introductory Physics Homework Help
Replies
23
Views
3K
• Introductory Physics Homework Help
Replies
2
Views
1K