Calculate Work Done by Gravity: Physics Homework Problem

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To calculate the work done by gravity on a 0.179 kg ball thrown upward from 1.94 m, the force of gravity is determined using F = mg, yielding 1.7562 N. The displacement is -1.94 m, indicating downward movement. The work done by gravity is calculated using W = Fd, resulting in a positive value since the force of gravity acts in the same direction as the displacement. Proper sign conventions are crucial, as both the force and displacement should be negative if considering upward motion as positive. The final calculation confirms that gravity does positive work on the ball as it falls.
reaperkid
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Homework Statement



A 0.179 kg ball is thrown straight up from 1.94 m above the ground. Its initial vertical speed is 10.00 m/s. A short time later, it hits the ground. Calculate the total work done by the force of gravity during that time.

m = .179 kg
vi = 10m/s (upward)
g = 9.8 m/s2
d = -1.94 m (since the displacement is downward)

Homework Equations



W = Fd
F = ma

The Attempt at a Solution



This seemed like a simple problem, I'm not sure why it's giving me so much trouble. Basically, I just need to find the force of gravity which I thought was just 9.8 * .179 but I'm clearly missing something.

I tried... W = -1.94 * (9.8 * .179)

Thanks
 
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reaperkid said:

Homework Statement



A 0.179 kg ball is thrown straight up from 1.94 m above the ground. Its initial vertical speed is 10.00 m/s. A short time later, it hits the ground. Calculate the total work done by the force of gravity during that time.

m = .179 kg
vi = 10m/s (upward)
g = 9.8 m/s2
d = -1.94 m (since the displacement is downward)

Homework Equations



W = Fd
F = ma

The Attempt at a Solution



This seemed like a simple problem, I'm not sure why it's giving me so much trouble. Basically, I just need to find the force of gravity which I thought was just 9.8 * .179 but I'm clearly missing something.

I tried... W = -1.94 * (9.8 * .179)

Thanks

x_f^2=x_i^2+2ad, W=Fd\rightarrow W=mg(x_f^2-x_i^2)/2a.
 
asleight said:
x_f^2=x_i^2+2ad, W=Fd\rightarrow W=mg(x_f^2-x_i^2)/2a.

That doesn't incorporate the initial velocity though does it?

Wouldn't

xi = 1.94 m ?
xf = 0 m ?
 
Ignore asleight's nonsense.

You did almost the correct thing: you chose the displacement as negative (fine, that's just a convention), but then you should take the acceleration due to gravity as negative, too! Both are downward.

And even without thinking about sign conventions: does gravity do positive or negative net work here?
 
I seee, it's positive because it's in the direction of gravity.

Thank you very much sir! :)
 
You're welcome!
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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