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.

a. If there were no friction, how fast would he be going when he reached the bottom of the hill? (1 point)

b. In reality, the friction between the sled and the ground causes Josh to lose 15% of his initial energy as heat. In this case, how fast will he be going when he reaches the bottom? (2 points)

c. Josh is sledding into a valley that rises again on the other side. He can sled down one side and then back up the other side. If he loses 15% of his initial energy to friction by the time he reaches the bottom, then loses another 10% of his remaining energy sledding back up, how high can he slide up the other side of the valley before he comes to a stop? (2 points)

d. Instead of starting from rest, Josh starts with a speed of 10 m/s. How fast will he be going when he reaches the bottom of the hill? (Assume no friction.) (2 points)

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:

[tex]K_i + U_i = K_f + U_f[/tex]

where

[tex]K_i = 0[/tex] is the kinetic energy on top (zero since Josh starts from rest)

[tex]U_i = mgh[/tex] 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

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

[tex]U_f = 0[/tex] is the gravitational potential energy at the bottom of the hill (zero since h=0)

So we can rewrite the equation as

[tex]mgh=\frac{1}{2}mv^2[/tex]

[tex]v=\sqrt{2gh}[/tex]

And using:

g = 9.8 m/s^2

h = 35 m

We find

[tex]v=\sqrt{2(9.8)(35)}=26.2 m/s[/tex]

b. 24.1 m/s

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

[tex]U_i = mgh = (60)(9.8)(35)=20580 J[/tex]

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

[tex]E_f = U_i - 0.15 U_i = 0.85 U_i = 0.85(20580)=17493 J[/tex]

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

[tex]K_f = E_f = 17493 J[/tex]

So we can find the new final speed:

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

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

[tex]E = K_1 - 0.10 K_1 =0.90 K_1 = 0.90 (17493)=15744 J[/tex]

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

[tex]U_2 = E = 15744 J[/tex]

So we have

[tex]U_2 = mgh_2[/tex]

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

[tex]h_2 = \frac{U_2}{mg}=\frac{15744}{(60)(9.8)}=26.8 m[/tex]

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:

[tex]K_i + U_i = K_f \rightarrow \frac{1}{2}mu^2+mgh=\frac{1}{2}mv^2[/tex]

where

u = 10 m/s is the initial speed

v is the final speed

Simplyfing the equation, we get

[tex]v=\sqrt{u^2+2gh}[/tex]

And using h = 35 m, we find

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