Question

A 5.75-kg block is sent up a ramp inclined at an angle θ = 32.5° from the horizontal. It is given an initial velocity v0 = 15.0 m/s up the ramp. Between the block and the ramp, the coefficient of kinetic friction is μk = 0.382 and the coefficient of static friction is μs = 0.687. How far up the ramp (in the direction along the ramp) does the block go before it comes to a stop?

___________m

Answer 1

By applying newton's law in inclined direction:

Fnet = m*a

-Fg*sin - Ff = m*a

-m*g*sin - k*N = m*a

here, N = normal force = m*g*cos

= angle = 32.5 deg

k = coefficient of kinetic friction = 0.382

So, a = -(g*sin + k*g*cos)

a = -(9.81*sin(32.5 deg) + 0.382*9.81*cos(32.5 deg))

a = -8.43 m/s^2 = acceleration of block on inclined ramp

now, using third kinematics law in inclined direction:

v^2 - u^2 = 2*a*S

here, v = final speed = 0

u = initial speed = 15.0 m/s

So, S = distance travelled on ramp = (v^2 - u^2)/(2*a)

S = (0 - 15^2)/(2*(-8.43))

S = 13.3 m