Answer:
Rolling case achieves greater height than sliding case
Step-by-step explanation:
For sliding ball:
- When balls slides up the ramp the kinetic energy is converted to gravitational potential energy.
- We have frictionless ramp, hence no loss due to friction.So the entire kinetic energy is converted into potential energy.
- The ball slides it only has translational kinetic energy as follows:
ΔK.E = ΔP.E
0.5*m*v^2 = m*g*h
h = 0.5v^2 / g
For rolling ball:
- Its the same as the previous case but only difference is that there are two forms of kinetic energy translational and rotational. Thus the energy balance is:
ΔK.E = ΔP.E
0.5*m*v^2 + 0.5*I*w^2 = m*g*h
- Where I: moment of inertia of spherical ball = 2/5 *m*r^2
w: Angular speed = v / r
0.5*m*v^2 + 0.2*m*v^2 = m*g*h
0.7v^2 = g*h
h = 0.7v^2 / g
- From both results we see that 0.7v^2/g for rolling case is greater than 0.5v^2/g sliding case.
I can't answer this question if we don't know by what scale the cylinder's radius was reduced. Luckily, I found the same problem that says the radius was reduced to 2/5. So, we find the ratio of both volumes.
V₁ = πr₁²h₁
V₂ = πr₂²h₂
where r₂ = 2/5*r₁ and h₂ = 4h₁
V₂/V₁ = π(2/5*r₁ )²(4h₁)/πr₁²h₁= 8/5 or 1.6
<em>Thus, the volume has increased more by 60%.</em>
$3 went to the men and and $2 to the bellboy. That's $5 in total, which is what the bellboy was supposed to give the men. So the $1 is with one of the men.
Rotational, horizontal, and vertical.
