Question

The block pictured below is sliding upward on the incline with theta=25. the coefficient of friction between the block and the incline is 0.15. of the block has a speed of 5 m/s at the bottom of the incline, how far upward will it slide before it stops?

I've got a few ideas on how to work this but I don't know how to integrate theta into the equation.

Answer 1

The sum of the forces in the direction parallel to the incline:

m*g*sin(theta) - cf*N = m*a
where m is the mass of the block, cf is the coefficient of friction, N is the normal force, and a is the acceleration

The sum of the forces in the direction perpendicular to the incline:

N - m*g*cos(theta) = 0 (because the block is not moving in the direction perpendicular to the incline)

N = m*g*cos(theta)

m*g*sin(theta) - cf*m*g*cos(theta) = m*a

g*sin(25) - 0.15*g*cos(25) = a

Now use

v = v0 - a*t
x = x0 + v0*t - .5*a*t^2

where v is speed, v0 is initial speed, t is time, x is distance, x0 is initial position
Plugging in numbers

0 = 5 - a*t
solve for t, using the answer for a you found above, then plug into the equation for x