Bowling Ball?

Question

A bowling ball encounters a 0.760-m vertical rise on the way back to the ball rack. Ignore frictional losses and assume that the mass of the ball is distributed uniformly. The translational speed of the ball is 7.83 m/s at the bottom of the rise. Find the translational speed at the top.

Answer 1

This is a conservation of energy problem

vf=speed at top
vi=speed at bottom 7.83 m/s
h=rise .76m
g=gravity 9.81 m/s^2

1/2*m*vi^2=1/2*m*vf^2+m*g*h

note that the mass divides out
vf=sqrt(vi^2-2*g*h)
=sqrt(7.83^2-2*9.81*.76)
=6.81 m/s

j

Answer 2

6.81 m/s

The amount of kinetic energy at the end is equaled to the kinetic energy at the beginning minus the potential energy gained as a result of the change in height.

So as an expression you would write:

KE(final) = KE (initial) - PE (final)

This assumes that the ball did not have any potential energy before it changes elevation.

0.5*m*v(final)^2 =(0.5*m*v(initial)^2) - mgh

The mass cancels out of each term, so you are left with

0.5*v(final)^2 =(0.5*v(initial)^2) - gh

Do the algebra to solve for v(final)

v(final) = sqrt(2(0.5*v(initial)^2) - gh))

Plug in what you know

v(final) = sqrt(2(0.5*(7.83^2))-(9.81*0.76)))

You should end up with

v(final) = 6.81 m/s

Answer 3

dont know. it sounds like a confusing process though.