# Physics Help Greatly Appreciated - Block sliding down ramp

Physics Help Greatly Appreciated -- Block sliding down ramp

A 7.7 kg block is on an incline with a coefficient of static friction of 0.50. The angle that the ramp makes with the horizontal is increased gradually, until the block begins to slide down the ramp. If you know that the coefficient of kinetic friction between the block and the plane is 0.34, find the time it will take the block to slide down the ramp a distance of 0.47 m starting from rest.

Ff=uFn

My solution:

arctan(.5)=angle of incline=26.6degrees
a=2x/t^2 (rearranged kinematic equation)
a=gsin(26.6)
a=4.39 m/s^2
4.39=2x/t^2
x=.47m
.94/4.39=t^2
t=.46 s

(This solution was marked incorrect) Help would be greatly appreciated.

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It seems to me that you have calculated the time it would take if the coefficient of kinetic friction were zero.

Vic is correct. Letting the acceleration = gsin(theta) gives the acceleration in the absence of friction. If you look at the force balance in the down incline direction it should read:

Ffriction + Fgravity = ma (1)

<< some extra work deleted by Mentor >>

Solve that to find the acceleration and the re-evaluate your kinematics equation.

Chris.

Last edited by a moderator:
berkeman
Mentor
Vic is correct. Letting the acceleration = gsin(theta) gives the acceleration in the absence of friction. If you look at the force balance in the down incline direction it should read:

Ffriction + Fgravity = ma (1)

<< some extra work deleted by Mentor >>

Solve that to find the acceleration and the re-evaluate your kinematics equation.

Chris.
Thanks for being willing and able to help with this question, Alexander. Just please remember that the OP student must do the bulk of the work. I think with the hints that he's been given now, he should be able to show us his work toward the correct solution.

Thanks for the help! I genuinely appreciate it :)
I used
Fg-Ff=ma
mgsin(26.6)-umgcos26.6)=ma
9.8(sin26.6)-.32cos(26.6)=a
a=1.41
t=.82s