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:
[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]
