Block and ramp help

Question

A block starts with a speed of 15 m/s at the bottom of a ramp that is inclined at an angle of 30 ̊ above the horizontal. The coefficient of kinetic friction between the block and the plan is 0.2. The block goes up the ramp, momentarily comes to rest, and then slides back down the ramp. What is the speed of the block when it reaches the bottom of the ramp?

Answer 1

First, find the net force (Fnet) on the ramp. As it goes up the ramp two forces act against its motion to produce negative acceleration: the gravity component parallel to the ramp (Fp) and the friction force (Ff).

Fp = weight (mg) times sin 30 = 0.5mg

Normal force (FN) is weight times cos 30 = 0.87mg

Ff is coefficient of kinetic friction (µ) times FN = (0.2)(0.87)mg = 0.17mg

Fnet = Fp + Ff = ma = 0.5mg + 0.17mg; divide both sides by m a = 0.5g + 0.17g = 0.67(-9.8) = -6.57m/s²

Now find the distance the block travels up the ramp before stopping: v²_f = v²_i + 2ad

0 = 15² + 2(-6.57)d; d = 17.1 m

Now find Fnet and acceleration down the ramp: Fnet = Fp - Ff = ma = 0.5mg - 0.17mg

a = 0.33(-9.8) = -3.23 m/s²

Find speed at the bottom of the ramp: v²_f = 0 + 2(-3.23)(-17.1) = 110.6

v_f = sqrt(110.6) = 10.5 m/s