Sliding down a ramp against friction force

[Vader07]

Homework Statement

At the top of the ramp the body is stopped and then released. It slides down the ramp against a friction force of f= 10 Newtons. Find the body's change in kinetic energy in Joules after traveling a distance s= 3 m down the ramp.

Homework Equations

There are three types of energy:

1) potential energy: Epot : the body is on top of the ramp (height h) at rest. The energy is Epot = mgh
2) frictional energy: Efric = cmg*cos alpha *s (= frictional force multiplied by the distance s that the body had gone down the ramp at a certain point.)
3) kinetic energy: Ekin = 1/2 mgv^2 : let's you calculate the velocity of the body.

The three energies are related in the following way:
Epot = Efric + Ekin, -->
Ekin = Epot - Efric .

The Attempt at a Solution

KE = 1/2mgv^2
KE = 1/2(1)(9.8)^2
= 1/2(96.04)
= 49.04

F1 = (1)(9.8)sin(alpha)

 

[dacruick]

the change in kinetic energy is simply going to be your change in potential energy minus the energy lost to friction(Or the work done by friction). With emphasis on work.

 

[Vader07]

dacruick


thanks for the reply. could you help me set the equation for this change in kinetic energy. I am sort of lost on this.

 

[dacruick]

just take the kinetic energy at the start and the kinetic energy at the end, and then find the difference. do you know the equation for kinetic energy?

 

[rwisz]

= (1)(9.8)sin(0) = 0
F2 = (1)(9.8)cos(alpha) = (1)(9.8)cos(0) = 9.8

Ffric = μFn = μmgcos(alpha) = (0.2)(1)(9.8)cos(0) = 1.96

Efric = Ffric * s = (1.96)(3) = 5.88

Thus, the change in kinetic energy is:

ΔKE = KEfinal - KEinitial = (1/2)(1)(vfinal)^2 - (1/2)(1)(vinitial)^2 = (1/2)(1)(0)^2 - (1/2)(1)(9.8)^2 = -48.02 Joules

The negative sign indicates that the body's kinetic energy has decreased due to the work done by the friction force.