Calculating Friction Force in Sliding Block Problem

[aburich_44]

A 38.5g block is sliding at 5.0m/s on the upper surface which is frictionless and 1.00m above the lower frictionless surface. The block slides down a ramp that is at an angle of 18.5 degrees. What must the friction force on the ramp be in order for the block to have a velocity of 5.5m/s at the bottom of the ramp?? I got the equation and an answer, but I am not sure if its right. how can i finish?

 

[LowlyPion]

A 38.5g block is sliding at 5.0m/s on the upper surface which is frictionless and 1.00m above the lower frictionless surface. The block slides down a ramp that is at an angle of 18.5 degrees. What must the friction force on the ramp be in order for the block to have a velocity of 5.5m/s at the bottom of the ramp?? I got the equation and an answer, but I am not sure if its right. how can i finish?

Welcome to PF.

What is your equation and where are you stuck?

 

[aburich_44]

i got PEg (gravitational) - w (since the direction of friction force is against the displacement) = kef . And then i have w= fd; so f= (kf-peg) over -d. so that's equal to (1/2mv(final) ^2 - mgh) all over -d. Now I am stuck ?

 

[LowlyPion]

OK.

How much energy will it gain from Potential Energy? m*g*h now I'm sure you can figure the h.
So how much of that made it to Kinetic energy if the velocity went from 5 to 5.5m/s?

So isn't it that whatever that difference is between energy added from Potential and what actually became Kinetic must have gone into Work from friction?

And over that distance that would be how much Force ...?

And this Force is given by ...?

Which means that μ will need to be ...?

 

[aburich_44]

hmm let's see: i did (1/2(0.0385)(5.5)^2 - 0.0385(9.81)(1) ) all over -3.15 (because h =1 and the distance is 1/sin18.5. So my answer is -0.65 ?? is that right?