Finding the Compression of a Spring Using the Work-Energy Theorem

AI Thread Summary
A 6.0kg box moving at 3.0m/s collides with a spring of force constant 75 N/cm, and the work-energy theorem is used to find the maximum compression of the spring. The initial calculations incorrectly used a spring constant of 75,000 N/m instead of the correct value of 7,500 N/m. This led to an initial compression result of 0.027m, while the textbook states the answer should be 0.085m. The error was identified as a misconversion from N/cm to N/m. Correcting this conversion resolves the discrepancy in the calculated maximum compression.
student34
Messages
639
Reaction score
21

Homework Statement



A 6.0kg box moving at 3.0m/s on a horizontal, frictionless surface runs into a light spring of force constant 75 N/cm. Use the work-energy theorem to find the maximum compression in the spring.

Homework Equations



W = K2 - K1

W = (1/2)*m*v^2 - (1/2)*m*vo^2

W = - (1/2)*k*x^2

The Attempt at a Solution



- (1/2)*k*x^2 = 0 - (1/2)*m*vo^2, where k = 75000N/m

x = ((m*v^2)/k)^(1/2) = ((6.0kg*(3m/s)^2)/75000N/m)^(1/2)

x = 0.027m

But the answer in the book is x = 0.085m. I just can't see what I am doing wrong unless my textbook is wrong.
 
Physics news on Phys.org
Recheck your value for k.
 
Averki said:
Recheck your value for k.

Oh my god, thanks!
 
student34 said:

Homework Statement



A 6.0kg box moving at 3.0m/s on a horizontal, frictionless surface runs into a light spring of force constant 75 N/cm. Use the work-energy theorem to find the maximum compression in the spring.

Homework Equations



W = K2 - K1

W = (1/2)*m*v^2 - (1/2)*m*vo^2

W = - (1/2)*k*x^2

The Attempt at a Solution



- (1/2)*k*x^2 = 0 - (1/2)*m*vo^2, where k = 75000N/m

x = ((m*v^2)/k)^(1/2) = ((6.0kg*(3m/s)^2)/75000N/m)^(1/2)

x = 0.027m

But the answer in the book is x = 0.085m. I just can't see what I am doing wrong unless my textbook is wrong.

It's your conversion of N/cm to N/m sir. It is only 7,500 instead of 75,000. But thanks so much for this
 
semaj nayr said:
It's your conversion of N/cm to N/m sir. It is only 7,500 instead of 75,000. But thanks so much for this
:welcome:

Note that this thread is nine years old.
 
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Back
Top