How Does Spring Compression Affect Sled Velocity at a Waterpark?

  • Thread starter Thread starter Aiko
  • Start date Start date
AI Thread Summary
The discussion focuses on calculating the sled's velocity at a waterpark when released from a compressed spring. For Part A, the correct velocity when the spring returns to its uncompressed length is determined to be 2.67 m/s, based on the work done by the spring. In Part B, there is confusion regarding the correct compression distance, with the original question specifying 0.215 m. The user attempts various calculations to find the sled's speed at this compression, ultimately arriving at a velocity of 2.12 m/s using the correct energy equations. Clarification on the compression distance and the application of energy conservation principles is essential for accurate results.
Aiko
Messages
8
Reaction score
0

Homework Statement


At a waterpark, sleds with riders are sent along a slippery, horizontal surface by the release of a large, compressed spring. The spring with a force constant k = 3600 N/m and negligible mass rests on the frictionless horizontal surface. One end is in contact with a stationary wall. A sled and rider with total mass 75.0 kg are pushed against the other end, compressing the spring 0.385 m. The sled is then released with zero initial velocity.

A. What is the sled's speed when the spring returns to its uncompressed length?
B. What is the sled's speed when the spring is still compressed 0.215 m?

Homework Equations


W=K2-K1
W=1/2kx^2-1/2kx^2
W=1/2kx^2

The Attempt at a Solution


I solved A by using W=1/2kx^2 to find the work done by the spring to be 266.81J. I then used that along with V1=0, k=3600 and m=75.0 to find V2 which was 2.67m/s. I got this part right.

Part B is where I am having some trouble.

My first try, I tried to do the same thing as above.
I found the work done by the spring at .210m:
W=1/2(3600)(.210)^2=83.205J

and then I used this in the W=K2-K1 with V1 being zero
83.205=1/2(75)(V2)^2-1/2(75)(0)^2

V2=1.49m/s <--- this answer was wrong.

My second try, I did the same thing except I used W=1/2kx^2-1/2kx^2, using X1=.210m and x2=.385m. With this, I found the work done to be W=183.6J

Using that, I plugged it into the W=KE2-KE1 to find V1:
183.6=1/2(75.0)(2.67)^2-1/2(75.0)(V1)
V1=1.50

Can someone show me what I am doing wrong?
 
Physics news on Phys.org
For Part B, is it .210m or .215m as in the original question you posted?
 
B. What is the sled's speed when the spring is still compressed 0.215 m?
Still compressed means, is it in addition to 0.385 m?
 
rl.bhat said:
B. What is the sled's speed when the spring is still compressed 0.215 m?
Still compressed means, is it in addition to 0.385 m?

I think the OP has understood it correctly insofar as it wants to know what the speed is as it passes .215m going to 0 detent.
 
LowlyPion said:
For Part B, is it .210m or .215m as in the original question you posted?

It's .215m, my bad.
 
Aiko said:

Homework Statement


At a waterpark, sleds with riders are sent along a slippery, horizontal surface by the release of a large, compressed spring. The spring with a force constant k = 3600 N/m and negligible mass rests on the frictionless horizontal surface. One end is in contact with a stationary wall. A sled and rider with total mass 75.0 kg are pushed against the other end, compressing the spring 0.385 m. The sled is then released with zero initial velocity.

A. What is the sled's speed when the spring returns to its uncompressed length?
B. What is the sled's speed when the spring is still compressed 0.215 m?

Homework Equations


W=K2-K1
W=1/2kx^2-1/2kx^2
W=1/2kx^2

The Attempt at a Solution


I solved A by using W=1/2kx^2 to find the work done by the spring to be 266.81J. I then used that along with V1=0, k=3600 and m=75.0 to find V2 which was 2.67m/s. I got this part right.

Part B is where I am having some trouble.

My first try, I tried to do the same thing as above.
I found the work done by the spring at .210m:
W=1/2(3600)(.210)^2=83.205J

and then I used this in the W=K2-K1 with V1 being zero
83.205=1/2(75)(V2)^2-1/2(75)(0)^2

V2=1.49m/s <--- this answer was wrong.

My second try, I did the same thing except I used W=1/2kx^2-1/2kx^2, using X1=.210m and x2=.385m. With this, I found the work done to be W=183.6J

Using that, I plugged it into the W=KE2-KE1 to find V1:
183.6=1/2(75.0)(2.67)^2-1/2(75.0)(V1)
V1=1.50

Can someone show me what I am doing wrong?

Does anyone know what I am doing wrong?
 
Spring Potential E(.385) - Spring Potential E(.215) = 1/2*m*v2

3600/2*((.385)2 - (.215)2) = 1/2*75*V2

V2 = 1800*(.102)*2/75

V = 2.12
 
Back
Top