Answer Check: A golf ball is hit and hits a bee. find the work done on the bee

joe426

1. Homework Statement 2. Homework Equations

F=ma
W= Fdcosθ

3. The Attempt at a Solution

I first found the acceleration:
t= d/v
t= 37m/28m/s = 1.32s
a = v/t
a = 28m/s / 1.32s
a= 21.21 m/ss

Second, the force:
Fclub - Fgsinθ = ma
Fclub = a +gsinθ
Fclub = 21.21m/ss + 9.8m/ss sin84
Fclub = 30.96N

Then I plugged in:
W=(30.96N)(37m)cos84
W=119.75J

I'm pretty sure this is correct. only thing I'm worried about really is the acceleration.

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frogjg2003

Using energy considerations
You have to use potential and kinetic energy conservation to solve this problem.

haruspex

Homework Helper
Gold Member
2018 Award
I first found the acceleration:
t= d/v
t= 37m/28m/s = 1.32s
a = v/t
a = 28m/s / 1.32s
a= 21.21 m/ss
That calculation is completely wrong. 28m/s is the initial speed of the ball, and it is at some angle to the vertical. The flight time to reach the bee would be 1.32s if the ball were travelling straight up at 28m/s and not slowed by gravity along the way.
What acceleration do you think this is calculating?
Second, the force:
Fclub - Fgsinθ = ma
Fclub = a +gsinθ
Fclub = 21.21m/ss + 9.8m/ss sin84
Fclub = 30.96N
It is impossible to calculate the force from the club since you do not know how long the impact lasted. You can calculate the impulse (momentum), but that doesn't help with this question.

joe426

You have to use potential and kinetic energy conservation to solve this problem.
So, since we can ignore resistance, the energy is conserved and the potential energy at the beginning equals the kinetic energy at the end.

PEi = KEf
mgy = 1/2mv2
v = sqrt(2gy)
v = sqrt(2 * 9.8m/s2 * 37m)
v= 26.93m/s

haruspex

Homework Helper
Gold Member
2018 Award
So, since we can ignore resistance, the energy is conserved and the potential energy at the beginning equals the kinetic energy at the end.

PEi = KEf
This isn't a stone being dropped from rest. Total energy is conserved here means PEi + KEi= PEf + KEf
What is KEi?

joe426

This isn't a stone being dropped from rest. Total energy is conserved here means PEi + KEi= PEf + KEf
What is KEi?
KEi = 0 because the initial velocity is 0. And would make 1/2mv2 = 0

And PEf = 0 because y is the vertical distance to move the object. we dont need to move the object because its already at its finaly position of 37m. If wrong on this and I should use 37m for y and solve for v, I end up with 0

frogjg2003

KEi = 0 because the initial velocity is 0. And would make 1/2mv2 = 0
The initial kinetic energy is not zero. The ball has some initial speed, namely 28 m/s.

And PEf = 0 because y is the vertical distance to move the object. we dont need to move the object because its already at its finaly position of 37m. If wrong on this and I should use 37m for y and solve for v, I end up with 0
It's usually more convenient to have the starting, not final position have position and potential energy 0. This means that the final position is at y=37m and a potential energy of mgy.

You know the initial potential and kinetic energies, and the final potential energy. The only thing you don't know is the final kinetic energy. Solve for this and then determine the speed.

joe426

The initial kinetic energy is not zero. The ball has some initial speed, namely 28 m/s.

It's usually more convenient to have the starting, not final position have position and potential energy 0. This means that the final position is at y=37m and a potential energy of mgy.

You know the initial potential and kinetic energies, and the final potential energy. The only thing you don't know is the final kinetic energy. Solve for this and then determine the speed.
Ok, I understand this and I've come up with:
1/2mvi2 = mgyf + 1/2mvf2
vf = sqrt [ vi2 - 2(gyf) ]
vf = 7.67m/s

frogjg2003

There you go. You've got it.

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