(a) Yes
Explanation: the slide is frictionless: this means that there are no non-conservative forces acting on the child, therefore the mechanical energy of the child is conserved.
The mechanical energy is sum of the kinetic energy (K) and the gravitational potential energy (U) of the child:
[tex]E=K+U=\frac{1}{2}mv^2+mgh=const.[/tex]
where m is the mass of the child, v is his speed, g is the gravitational acceleration, h is his heigth above the ground.
(b) No
At the beginning of the motion of the child (highest point of the slide), the child is at rest, so its speed is zero: v=0, and all the mechanical energy is just potential energy:
[tex]E=U=mgh[/tex]
where h is the height of the slide.
At the bottom of the slide, the height is zero: h=0, so all the mechanical energy has been converted into kinetic energy:
[tex]E=K=\frac{1}{2}mv^2[/tex]
where v is the speed at the bottom of the slide. Since E is conserved, we can equalize the two equations, and we find an expression for v:
[tex]\frac{1}{2}mv^2=mgh\\v=\sqrt{2gh}[/tex]
And we see that v does not depend on the mass of the child.
(c) same speed in either case
We can repeat the same procedure used in the previous part of the exercise, by using the conservation of energy, and we see again that the final speed of the child at ground level depends only on the height of the slide, h:
[tex]v=\sqrt{2gh}[/tex]
Therefore, the child will be travelling at same speed.
(d) [tex]mgh = \frac{1}{2}mv^2 + E_{wasted}[/tex]
If friction is present, the mechanical energy is no longer conserved, because part of the energy has been wasted because of the friction. The equation of conservation of energy would become:
[tex]mgh = \frac{1}{2}mv^2 + E_{wasted}[/tex]
which means that the initial potential energy at the top of the ramp (mgh) has been converted only partially into kinetic energy ([tex]\frac{1}{2}mv^2[/tex]), while part has been wasted because of the frictional force ([tex]E_{wasted}[/tex]).
(e) 15.0 m/s
If we apply the formula found at point b), by using h=11.5 m, we can immediately find the speed of the child at the bottom in this situation:
[tex]v=\sqrt{2gh}=\sqrt{2(9.8 m/s^2)(11.5 m)}=15.0 m/s[/tex]
