Calculate Work Done by Gravity: Physics Homework Problem

In summary, the total work done by the force of gravity on a 0.179 kg ball thrown straight up from 1.94 m above the ground with an initial vertical speed of 10 m/s and hitting the ground is positive and equal to 1.12 J. This is calculated using the formula W = mg(x_f^2 - x_i^2)/2a, with displacement and acceleration both taken as negative due to the downward direction of gravity.
  • #1
reaperkid
14
0

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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  • #2
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

[tex]x_f^2=x_i^2+2ad[/tex], [tex]W=Fd\rightarrow W=mg(x_f^2-x_i^2)/2a[/tex].
 
  • #3
asleight said:
[tex]x_f^2=x_i^2+2ad[/tex], [tex]W=Fd\rightarrow W=mg(x_f^2-x_i^2)/2a[/tex].

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

Wouldn't

xi = 1.94 m ?
xf = 0 m ?
 
  • #4
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?
 
  • #5
I seee, it's positive because it's in the direction of gravity.

Thank you very much sir! :)
 
  • #6
You're welcome!
 

1. How do you calculate the work done by gravity?

The work done by gravity can be calculated by multiplying the force of gravity (weight) by the displacement (change in position) of an object in the direction of the force. The formula is W = mgh, where m is the mass of the object, g is the acceleration due to gravity (usually 9.8 m/s^2), and h is the vertical displacement.

2. What are the units of work done by gravity?

The units of work done by gravity are joules (J) in the SI system. In other systems, such as the imperial system, the units can be foot-pounds (ft-lb) or British thermal units (BTU).

3. Can the work done by gravity be negative?

Yes, the work done by gravity can be negative if the displacement of the object is in the opposite direction of the force of gravity. This can happen when an object is moving upwards against the force of gravity, or when the displacement is in a horizontal direction.

4. How does the mass of an object affect the work done by gravity?

The mass of an object directly affects the work done by gravity. The greater the mass, the greater the work done by gravity. This is because the weight (force of gravity) is directly proportional to the mass of an object according to Newton's second law of motion (F=ma).

5. Can the work done by gravity be used to find the speed of an object?

No, the work done by gravity does not directly give the speed of an object. However, it can be used in combination with other equations, such as the work-energy theorem, to calculate the speed of an object at a certain point in its motion.

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