Calculate speed of block at the end of the ramp

[Anon2459]

Homework Statement

A mass of m = 1.40 kg is placed on a 2.00 m ramp. The angle of the ramp can be adjusted by changing the height of the top of the ramp.

In reality there is a small amount of friction between the block and the ramp: μs = 0.261, μk = 0.119.

If the ramp (with friction acting) is now lifted so that θ = 48.0o and the block now slides from rest the full 2.00 m down the ramp what is its speed at the bottom?

Homework Equations

V^2 = -2 x g x (h-ugcosangle x d)

height of ramp = 0.7020408163

The Attempt at a Solution

V^2 = -2 x 9.8 x (-0.7020408163 + 0.119 x cos48 x 2) = wrong answer

Which friction do i use? Static or kinetic? Or is my approach wrong entirely?

Not sure where i went wrong[/B]

 

[Orodruin]

When you post, please provide your argumentation along with your equations and do not forget the units - units are important. Also, do not overdo the significant digits.

That being said: What is the work done by a force? In which direction does friction act? Consequently, what is the work done by the friction?

 

[dean barry]

Find out first if the force due to gravity can overcome the static friction force. if so it will accelerate down the block.

 

[haruspex]

height of ramp = 0.7020408163

It must be more than that.

 

[Anon2459]

hi everyone,

do i use us or uk to calculate the velocity?

Thanks

 

[haruspex]

hi everyone,

do i use us or uk to calculate the velocity?

Thanks

You are ignoring my post #4. Please show how you are calculating the ramp height.

 

[Anon2459]

You are ignoring my post #4. Please show how you are calculating the ramp height.

hi, sorry
i recalculated it using the new angle 48
so sin(48) x 2 = h

i incorrectly used the height from the previous answer hence why it was 0.7,

 

[Anon2459]

You are ignoring my post #4. Please show how you are calculating the ramp height.

My solution is:

V^2 = -2 x g x (-1.486289651 + 0.261 x cos(48) x 2)
And then square root

Is this correct ?

 

[haruspex]

My solution is:

V^2 = -2 x g x (-1.486289651 + 0.261 x cos(48) x 2)
And then square root

Is this correct ?

Looks ok.

 

[Anon2459]

Looks ok.

it was wrong unfortunately

 

[haruspex]

it was wrong unfortunately

What number did you get? I got roughly 5m/s.