A question regarding power and hookes law

Thread starter physicskid72
Start date May 5, 2010
Tags

Hookes law Law Power

AI Thread Summary

To determine the power required to compress a spring, the potential energy formula Pe = 1/2 kx^2 is used, where k is the spring constant and x is the compression distance. For a spring with a constant of 12 N/m compressed by 1.2 m, the potential energy calculated is 8.64 J. The power is then found by dividing the work done (8.64 J) by the time taken (2.5 s), resulting in 3.456 watts. The calculations presented appear to be correct, confirming the power requirement for the spring compression. The solution effectively demonstrates the application of Hooke's Law in calculating power.

May 5, 2010

physicskid72

Homework Statement

A 2.8 m high spring has a spring constant of 12 N/m. How much power is required if the spring is compressed 1.2 m in 2.5s?

Homework Equations

Pe= 1/2kx^2

Power= W/t

The Attempt at a Solution

=1/2(12)(1.2)^2
=8.64 J

so... Power = 8.64/2.5

= 3.456 watts

is this correct? thanks in advance!

Physics news on Phys.org

May 5, 2010

Doc Al

Mentor

Looks good to me.

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...

View full post »

Thread 'Explain this asphalt sheen'

This problem is two parts. The first is to determine what effects are being provided by each of the elements - 1) the window panes; 2) the asphalt surface. My answer to that is The second part of the problem is exactly why you get this affect. And one more spoiler:

View full post »