Question

MEDAL AND FAN
A 10 kg bowling ball sits at the top of a 10 m hill and then slides down its icy hillside.
a) What is the speed of the bowling ball when it reaches the bottom of the hill?
b) What is the change in kinetic energy of the system as the bowling ball travels from the top of the hill to the bottom of the hill?
c) What is the bowling ball’s mechanical energy at the top of the hill?
d) What is the bowling ball’s mechanical energy at the bottom of the hill?

Answer 1

For A:
Work=Force x Displacement
Force= Force of gravity = mass times acceleration due to gravity
Kinetic energy= 1/2 mass times speed

Using Work energy theorem
Work = Energy
F x d = 1/2 mv^2
m x g x d = 1/2 mv^2
g x d = 1/2 v^2

evaluate for v:
\(\sf v= \sqrt{2g \times d}\)

Answer 2

v=14.14m/s

Answer 3

yes v= 14 m/s

For B:
change in kinetic energy is equal to the final kinetic energy since initial kinetic energy is zero.. calculate using the speed calculated from A
Ek= 1/2mv^2

Answer 4

For C:
Mechanical Energy= Potential + Kinetic
mechanical energy at the TOP of the hill, so it means Ek is equal to zero,
calculate Ep using the formula Ep= mgh

Answer 5

Ek=980J

Answer 6

yes right (:

Answer 7

For D,
Mechanical Energy= Potential + Kinetic
mechanical energy at the bottom of the hill, this time Ep would be zero because h=0
therefore Mechanical energy is equal to Kinetic energy = 1/2mv^2 (which you already calculated in part B) (:

Answer 8

For C, ME=1000J?

Answer 9

Em= mgh + 0
= (10 kg)(9.8 m/s^2)(10m)
it's not equal to 1000J

Answer 10

oh nvm if you round 9.8 to 10 it will be the same

Answer 11

Thank you!

Answer 12

you're welcome :)