Question

An object of mass m is at rest at the top of a smooth slope of height h and length L. The coefficient of kinetic friction between the object and the surface, ?k, is small enough that the object will slide down the slope if given a very small push to get it started.
1. Find an expression for the object's speed at the bottom of the slope.
2. Sam, whose mass is 70kg , stands at the top of a 10-m-high, 80-m-long snow-covered slope. His skis have a coefficient of kinetic friction on snow of 0.07. If he uses his poles to get started, then glides down, what is his speed at the bottom?

Answer 1

Answer:

1.$v=\sqrt{2*h*g-2*u_K*g*L*cos*(arcsin(\frac{h}{L} ))}$

2. $v=9.33m/s$

Explanation:

1.

Kinetic energy and potential energy describe the motion also the small push means that initial kinetic energy is considered zero so:

Work done=Potential energy-Kinetic energy

$W_k=E_p-E_k$

$u_K*m*g*d=m*g*h-\frac{1}{2}*m*v^2$

$d=L*cos*(arcsin(\frac{h}{L} ))$

$u_K*g*L*cos*(arcsin(\frac{h}{L} ))=g*h-\frac{1}{2}*v^2$

Solve to v

$v^2=2*h*g-2*u_K*g*L*cos*(arcsin(\frac{h}{L} ))$

$v=\sqrt{2*h*g-2*u_K*g*L*cos*(arcsin(\frac{h}{L} ))}$

2.

Replacing the given: to find speed at the bottom

$v=\sqrt{2*10m*9.8m/s^2-2*0.07*9.8m/s^2*80m*cos(arcsin(\frac{10m}{80m}))}$

$v=\sqrt{87.1m^2/s^2}$

$v=9.33m/s$