Question

A 60kg skateboarder starts up a 20° slope at 5m/s then falls and slides up the hill on his knee pads. The coefficient of kinetic friction between his knee pads and the ramp is 0.30 . How far does he slide befor stopping?

Answer 1

2.0 meters  The skateboarder has 2 forces acting upon him to slow him down. The forces are friction, and climbing against the gravitational acceleration. So let's calculate the magnitude of these forces to see how fast he's decelerated.  The coefficient of kinetic friction is a multiplier to use against the normal force of the object. We can calculate the normal force by multiplying the mass of the object by the local gravitational acceleration and the cosine of the angle. So Df = 60 kg * 9.8 m/s^2 * cos(20°) * 0.30 Df = 60 kg * 9.8 m/s^2 * 0.939692621 * 0.30 Df = 60 kg * 9.8 m/s^2 * 0.939692621 * 0.30 Df = 165.7617783 kg*m/s^2 Df = 165.7617783 N  
The second amount of force is that caused by gravitational acceleration while climbing. That is determine by the amount of height gained for every meter along the slope. We can calculate that using the sine of the angle. So 
Dg = 60 kg * 9.8 m/s^2 * sin(20°)
 Dg = 60 kg * 9.8 m/s^2 * 0.342020143
 Dg = 201.1078443 kg*m/s^2
 Dg = 201.1078443 N
 
 So the amount of force decelerating the skateboarder is:
 F = Df + Dg
 F = 165.7617783 N + 201.1078443 N
 F = 366.8696226 N
 
 Now let's determine how much kinetic energy needs to be dissipated. The equation is
 E = 0.5 MV^2 
 So we'll substitute the known values and calculate
 E = 0.5 MV^2
 E = 0.5* 60 kg * (5 m/s)^2
 E = 0.5* 60 kg * 25 m^2/s^2
 E = 750 kg*m^2/s^2
 E = 750 J 
 Now let's divide the energy by the force.
 750 kg*m^2/s^2 / 366.8696226 kg*m/s^2 = 2.04432298 m
  Rounding to 2 significant figures gives a distance of 2.0 meters.