Help with physics work?

Question

A 6.7 kg object is released from rest at a height of 2.5 m on a curved frictionless ramp. At the foot of the ramp is a spring of force constant 225N/m. The object slides down the ramp and into the spring, compressing it a distance of x before coming momentarily to rest. The acceleration of gravity if 9.81 m/s^2. Find x.

I know this question is pretty long & complicated, but I'm having a test on problems like this soon, so I reeeeally need to know how to do it..and as of now...i have no clue. so can anyone help me out? please?

Answer 1

Lol, Ross M, kd is the force, not the energy. Elastic potential energy is (1/2) kd^2. So, you would get the wrong answer using that equation.

Also, kdruse, you can't use force to answer the question.

Answer 2

http://farm1.static.flickr.com/145/37743...

Answer 3

So, basically, this is an energy problem. It's actually pretty simple, but I don't have a calculator with me at the moment, so I'll give you the guide to solving it.

Since the object comes to a rest at the end of the problem the initial and final kinetic energies are both 0. Therefore, all the gravitational potential energy present at the beginning of the problem is converted completely into elastic potential energy (the spring) at the end of the problem. So, Ug = Us.

Ug = mgy

Us = (1/2) k (x^2)

mgy = (1/2) k (x^2)

Solving for x, we get:

x = square root (2mgy/k)

Plug in the numbers and get the answer. (m = 6.7 kg, g = 9.81 m/s^2, y = 2.5 m, and k = 225 N/m)

Answer 4

You might use the formula F=-kx, where F is the force of impact, k is the spring constant (or spring force constant), and x is the distance the spring compresses.

Answer 5

This question is about different types of energy.

At the top of the ramp the object has a certain Potential Eneregy, PE.

PE = mgh (mass x acc-of-gravity x height)

As is slides down the ramp it looses PE but gains Kinetic Energy.

In this problem you don't need to do anything with it.

BUT, the KE goes into compressing the spring.

At the instant that the object is stationary (at rest), all the KE is stored in the compressed spring as PE again.

PE (spring) = kd (force constant x distance compressed)

So, mgh = kd

You know m, g, h and k so you can calculate what d is. Or x in this case.

I hope this helps.  

Answer 6

Use the law of conservation of energy in this problem.

The potential energy of the object (since it is sliding down a frictionless ramp) will all be absorbed the the spring.

Mathematically,

Potential energy = mgh

Energy absorbed by spring = (1/2)kx^2

where

m = mass of the body = 6.7 kg (given)

g = acceleration due to gravity = 9.81 m/sec^2 (constant)

h = 2.5 m (given)

k = spring constant = 225 N/m

x = spring deflection

Substituting appropriate values,

(6.7)(9.81)(2.5) = (1/2)(225)(x^2)

Rearranging,

x^2 = (2)(6.7)(9.81)(2.5)/225

x^2 = 1.4606

x = 1.21 m

NOTE: I just cannot believe that a spring can be compressed this much!!!

Answer 7

physics is stupid, this question makes no sense where are you going to find a frictionless ramp?

Answer 8

OK here's how:

The height is potential energy. Falling down the ramp converts that to kinetic energy.  Calculate the potential energy.  That gives you the velocity of the body at the bottom (same as if it just dropped from a height because of the frictionless specification), but you probably don't need to calculate it.  The compression of the spring is going to be to the point that the potential energy of the spring is equal to the initial potential energy of height of the mass.  Why?  The mass is at the height of the spring, and is not moving, and energy didn't go anywhere but into the spring.  So, you have the equations for potential energy of a mass in gravity, and of a compressed spring.  Set them equal for this case, and then solve for the distance X.