Finding distance traveled up a ramp given m, angle, and velocity

[emilea]

Homework Statement

A block with mass m=18.4 kg slides down a ramp of slope angled 43.5 deg with a constant velocity. It is then projected up the plane with a Vo=1.55 m/s. How far does it go before stipping?

Homework Equations

Fnet=ma
Fg + Ff=Fnet
Ff=uk(Fn)
Fn=mgcos(theta)
a=(uk)cos(theta) + gsin(theta)

The Attempt at a Solution

I drew a free body diagram to solve for the coefficient of friction down the ramp, which was uk=0.948. I then used that to find the acceleration down the ramp using the a=(uk)cos(theta) + gsin(theta), which was 7.45. Here is where I got stuck. I think I could use deltax=Vo(t) + (1/2)a(t)^2, but I don't know if the downward acceleration would still be applicable or how to find the time. The equation I was thinking of using is t=x/(Vocostheta)), but I don't know x or t. could I combine that equation with x=Vo(t) + 1/2a(t^2)? Am I even on the right track?

 

[kuruman]

I drew a free body diagram to solve for the coefficient of friction down the ramp, which was uk=0.948. I then used that to find the acceleration down the ramp using the a=(uk)cos(theta) + gsin(theta), which was 7.45.

If the block slides down the incline at constant velocity, the acceleration must be ... ? Drawing a free body diagram is the way to go. If you still cannot finish the problem, please post the details of what you did so that we can correct anything that might be wrong.

 

[emilea]

I figured out that the downward acceleration is zero becausen it is at a constant velocity, which means the net downward force is also zero. I recalculated uk as 0.83 by setting Fg + Ff=0. My problem now is I have no idea how to connect this to any equations I could use to solve for x. I tried just using an equation for stopping distance: d=(Vo^2)/2ug, but that didn't work.

 

[kuruman]

I think your initial coefficient of kinetic friction (0.948) is the correct one. I questioned your statement that the acceleration down is not zero. Now you need to find the acceleration of the block when is is sliding up the incline. You already have an expression for that, a=(uk)cos(theta) + gsin(theta) in a direction down the incline. To find how far up the incline the block goes before it stops, use the kinematic equation that involves speed and displacement, but not time.

 

[jhae2.718]

\mu_k=0.948 is what I get.

If I'm understanding you correctly, emilea, I believe the expression you've posted for the acceleration down the plane [a=(uk)cos(theta) + gsin(theta)] is incorrect. Keep in mind that friction wants to act opposite the direction of motion.