When the ice cube is released, what distance will it travel up the slope before reversing direction?

Question

A 53.0 g ice cube can slide without friction up and down a 35.0 degree slope. The ice cube is pressed against a spring at the bottom of the slope, compressing the spring 12.0 cm. The spring constant is 23.0 N/m.

p.s. it is frictionless

Answer 1

Since no energy is lost to friction or any other forces, this problem can be solved using energy methods. The energy in the compressed spring is simply:

Es = 1/2 * k * x²

As the spring expands, it does work on the ice cube to speed it up. However, not all of the energy in the compressed spring is converted into kinetic energy of the ice cube. Note that gravity will pull the ice cube down the ramp according to the equation:

F = m*g*sin(theta)

The energy imparted from the spring to the cube will be equal to the spring energy MINUS the work done against gravity. YOU HAVE TO TAKE INTO ACCOUNT THE NEGATIVE WORK DONE BY GRAVITY! This work is:

W = F * x

W = m*g*sin(theta) * x

Therefore, the net energy imparted to the ice cube by the spring is simply:

KE = Es - W

KE = 1/2 * k * x² - m*g*sin(theta) * x

This is the kinetic energy of the cube after leaving the spring. Since no other forces are present, once the ice cube reaches its maximum height, its potential energy will be equal to its initial kinetic energy:

PE = KE = 1/2 * k * x² - m*g*sin(theta) * x

Potential energy is simply:

PE = m*g*h

m*g*h = 1/2 * k * x² - m*g*sin(theta) * x

whereh is the height of the block above the surface. Using simple trigonometry:

h = L sin (theta)

where L is the distance the ice cube travels up the slope after leaving the spring. Substituting this expression into the potential energy relation provides:

m*g*L sin (theta) = 1/2 * k * x² - m*g*sin(theta) * x

Solving for L:

L = (1/2 * k * x² - m*g*sin(theta) * x) / (m*g*sin(theta))

Finally, substituting all of the known values yields the answer:

L = [(.5)(23 N/m)(.12 m)² - (.053 kg)(.12 m)(9.81 m/s²)sin(35°)] / [(.053 kg)(9.81 m/s²)sin(35°)]

L = .4353 m

The ice cube will travel 43.53 cm up the ramp after leaving the spring. Hope that makes it clear! :)

Answer 2

Wow, that would make a GREAT homework question...


Most likely, the cube would succumb to differential drag on its non-aerodynamic surface, veer to one side and probably exit the ramp to one side or the other.

We also need to know such things as density of the atmosphere, temperature (is the device located in the heliosphere of a star for instance?).

And of course, how long is the slope?

Answer 3

Energy,E=T1+V1=T2+V2 where T=Kinetic energy and V=potential energy at state 1 and 2.

At state-2, when it reverses, velocity will be zero momentarilly. so,T2=0
T1=0

V1=0.5*k*x^2=mgh=V2, this is the eqn to solve

where, k=23N/m, x=12cm, m=53g , h=L *sin35
L=required distance it will climb up the hill.

Plug in values.
you will find distance =

I don't have time to grab calculator to solve yahoo answer question. hope you understand why!

Answer 4

nothing is frictionless unless this is fiction