# Work and energy

runningirl

## Homework Statement

Students are playing with the 4.13 kg bowling ball hung from the ceiling of the classroom. When at rest, it hangs 0.63 m above the floor. The cord is 1.87 long. The students pull the ball up and to the side and then let go. It swings back and forth like a pendulum. The students pull it so it is 1.40 m above the ground.

a) What is the ball's velocity when it's at the lowest point? Assume no air resistance.

b) What's the tension in the string at this point?

## The Attempt at a Solution

a) (1/2)mv^2=mgh
.5(4.13)v^2=4.13*9.8*.63
v=3.51 m/s?
b) mgh=F(1.4-.63)
9.8*4.13*.63=F(.77)
F=33.11 N?

I don't know if I did a right, and I don't know how to do b...

Last edited:

Mentor
a) (1/2)mv^2=mgh
.5(4.13)v^2=4.13*9.8*.63
v=3.51 m/s?
That's the correct formula, but h represents the change in height. From the highest point to the lowest point, what is the change in height of the ball?

For b you need to use Newton's 2nd law and some facts about circular motion.

runningirl
um... exactly what would i need for b?
"facts about circular motion" i mean.

Thanks for the help with part a!

runningirl
um... exactly what would i need for b?
"facts about circular motion" i mean.

Thanks for the help with part a!

runningirl
Oh. I get it now.

a=v^2/r
F=ma
Just find F.

Thanks!

Mentor
a=v^2/r
F=ma
Just find F.
Almost. F is the net force. There are two forces acting on the ball. One is the tension, which is what you need to find.

runningirl
Fnet=mg+tension?
but mg=-9.8(4.13)
so i could easily find tension?

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