Question

Josh sleds down a hill from a height of 35 m. He has a mass of 60 kg. He starts from rest at the top of the hill.

Answer 1

a. 26.2 m/s

Since there is no friction, we can apply the law of conservation of energy. The total mechanical energy (sum of potential energy + kinetic energy) must remain constant during the motion. So we can write:

$K_i + U_i = K_f + U_f$

where

$K_i = 0$ is the kinetic energy on top (zero since Josh starts from rest)

$U_i = mgh$ is the gravitational potential on top (measured relative to the bottom of the hill), with m being Josh's mass, g the acceleration of gravity, h the heigth of the hill

$K_f = \frac{1}{2}mv^2$ is the kinetic energy on the bottom of the hill, with v being Josh's final speed

$U_f = 0$ is the gravitational potential energy at the bottom of the hill (zero since h=0)

So we can rewrite the equation as

$mgh=\frac{1}{2}mv^2$

$v=\sqrt{2gh}$

And using:

g = 9.8 m/s^2

h = 35 m

We find

$v=\sqrt{2(9.8)(35)}=26.2 m/s$

b. 24.1 m/s

The initial energy that Josh has is the gravitational potential energy at the top of the hill:

$U_i = mgh = (60)(9.8)(35)=20580 J$

Josh loses 15% of this energy as heat, so the amount of mechanical energy left at the bottom of the hill is

$E_f = U_i - 0.15 U_i = 0.85 U_i = 0.85(20580)=17493 J$

This energy is converted into kinetic energy at the bottom of the hill:

$K_f = E_f = 17493 J$

So we can find the new final speed:

$K_f = \frac{1}{2}mv^2 \rightarrow v=\sqrt{\frac{2K_f}{m}}=\sqrt{\frac{2(17493)}{60}}=24.1 m/s$

c. 26.8 m

The total energy that Josh has at the bottom of the first hill is 17 493 J. Then he loses another 10% of its energy when going up the second hill: so the total energy at the top of the second hill is

$E = K_1 - 0.10 K_1 =0.90 K_1 = 0.90 (17493)=15744 J$

This energy is converted into gravitational potential energy at the top of the second hill:

$U_2 = E = 15744 J$

So we have

$U_2 = mgh_2$

and from this we can find h2, the maximum height that Josh can reach on the second hill:

$h_2 = \frac{U_2}{mg}=\frac{15744}{(60)(9.8)}=26.8 m$

d. 28.0 m/s

In this case, Josh does not start from rest, so its initial kinetic energy is not zero. So the equation of conservation of energy becomes:

$K_i + U_i = K_f \rightarrow \frac{1}{2}mu^2+mgh=\frac{1}{2}mv^2$

where

u = 10 m/s is the initial speed

v is the final speed

Simplyfing the equation, we get

$v=\sqrt{u^2+2gh}$

And using h = 35 m, we find

$v=\sqrt{(10)^2 + 2(9.8)(35)}=28.0 m/s$