Spring/Velocity problem

  • Thread starter Aiko
  • Start date
In summary, the question asks for the speed of a sled and rider on a slippery surface, propelled by a compressed spring with a force constant of 3600 N/m and negligible mass. The sled with a total mass of 75.0 kg is pushed against the spring, compressing it by 0.385 m, and then released with zero initial velocity. Part A asks for the speed when the spring returns to its uncompressed length, and Part B asks for the speed when the spring is still compressed at 0.215 m. To solve for Part B, the equation W=1/2kx^2-1/2kx^2 should be used, and the correct answer is 2.12
  • #1
Aiko
8
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?
 
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  • #2
For Part B, is it .210m or .215m as in the original question you posted?
 
  • #3
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?
 
  • #4
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.
 
  • #5
LowlyPion said:
For Part B, is it .210m or .215m as in the original question you posted?

It's .215m, my bad.
 
  • #6
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?
 
  • #7
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
 

1. What is a spring/velocity problem?

A spring/velocity problem is a type of physics problem that involves determining the displacement, velocity, acceleration, or other parameters of an object attached to a spring. The problem is usually presented in the form of a scenario where the spring is either compressed or stretched and then released, causing the object attached to it to move.

2. What are the key equations used to solve a spring/velocity problem?

The key equations used to solve a spring/velocity problem are Hooke's Law and the equations for simple harmonic motion. Hooke's Law states that the force exerted by a spring is directly proportional to the displacement of the object attached to it. The equations for simple harmonic motion describe the motion of the object in terms of its displacement, velocity, and acceleration.

3. How do you determine the spring constant in a spring/velocity problem?

The spring constant can be determined by dividing the force applied to the spring by the displacement of the object attached to it. This value remains constant for a given spring and is usually measured in units of newtons per meter (N/m).

4. How does the mass of the object affect the motion in a spring/velocity problem?

The mass of the object affects the motion in a spring/velocity problem by changing the natural frequency of the system. A heavier object will have a lower natural frequency, meaning it will take longer to complete one full cycle of motion compared to a lighter object. This can also affect the amplitude and velocity of the object's motion.

5. What are the units of measurement used in a spring/velocity problem?

The units of measurement used in a spring/velocity problem will depend on the specific parameters being solved for. Common units used include meters (m) for displacement, meters per second (m/s) for velocity, meters per second squared (m/s^2) for acceleration, newtons (N) for force, and kilograms (kg) for mass.

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