Block of ice sliding down inclined plane - final speed

[Edwardo_Elric]

Homework Statement

A block of ice with mass 2.00kg slides 0.70m down an inclined plane that slopes downward at an angle of 30degrees below the horizontal. If the block of ice starts from rest, what is its final speed?

Homework Equations

W = 1/2(m)(v2)^2 - 1/2(m)(v1)^2

The Attempt at a Solution

W = 1/2(2.00kg)(v2)^2 - 1/2(2.00kg)(0)^2

Fd = 1/2(2.00kg)(v2)^2

wsin(theta) is the only force horizontaly:
(wsin(theta))(0.70m) = 1/2(2.00kg)(v2)^2
2 * ((2.00kg)(9.80m/s^2))(0.70)) = 2.00kg (v2)^2

v2 = 3.70m/s

 

[mgb_phys]

Ignore the forces and look at the energy.
Work out the vertical distance from the length and the angle
You can then work out the potential energy lost PE = mgh
This is equal to the kinetic energy at the end KE=1/2 mv^2

 

[Edwardo_Elric]

but this lesson hasnt reach PE yet... this is still on work & kinetic energy
thanks btw

 

[learningphysics]

You forgot the sin(30), in your final line of calculations. It should be:

2 * ((2.00kg)(9.80m/s^2)sin(30))(0.70)) = 2.00kg (v2)^2