Conservation of Energy: work done by air resistance on rock

[Honey Bee]

Homework Statement

A 250 gram rock is thrown with a speed of 30.0 m/s. It has a speed of 22.5 m/s just before it strikes the ground. Determine the work done by air resistance.

There is also a diagram which yields more information. The rocks path is horizontal and it strikes the corner of a box that is 4.60 meters tall, ricocheting the rock at a 25 degree angle. (That is, the ball is projected at a 25 degree angle from a height of 4.6 meters).

Homework Equations

Ei+Pi=Kf+Pf+(work done by air resistance)

The Attempt at a Solution

Kf=63.3J, Ki=113J Pf=0, Pi=?

I am having difficulty with the idea of the horizontal path the rock travels and the gravitational potential energy... As well as what exactly to do with the the angle of projection. I understand that I need the initial and final conditions only... but this one is really stumping me. I thank you all in advance for your considerations and help.

 

[collinsmark]

Hello Honey Bee,

Welcome to Physics Forums!

I'm having trouble understanding the path. The problem statement says the rock strikes the ground, but then you say it strikes the corner of a box? Does the rock launched from a height of 4.60 m off the ground or does it land at an elevation 4.60 m off the ground, or both? Or does this collision with the box happen somewhere in between the time it is launched until the time it hits the ground?

Is there any way you could attach that figure? Are you sure the figure is for the same problem? (The problem statement and your description of the figure don't seem to match.)

 

[paisiello2]

Can we assume the collision with the box is perfectly elastic?

 

[Honey Bee]

Ah I am sorry. So as the diagram appears, The rock is thrown 4.6 meters above the ground and is traveling horizontally. It then collides with the corner of the box and ricochets off at a 25 degree angle, taking a parabolic path to the ground.

I am not sure about the collision. There is a decrease in its velocity from when it is thrown to just before it strikes the ground, so I imagine it is not perfectly elastic. But that is all the information given.

Here is the diagram. I am a newb and did not realize I could do this or I would have on my original post.

 

[paisiello2]

I guess you have to neglect the fact that nothing can be thrown that travels perfectly horizontally or just neglect this portion of the work done by air friction.

I would recommend to break the problem into two components: x and y directions.

 

[collinsmark]

Ah I am sorry. So as the diagram appears, The rock is thrown 4.6 meters above the ground and is traveling horizontally. It then collides with the corner of the box and ricochets off at a 25 degree angle, taking a parabolic path to the ground.

I am not sure about the collision. There is a decrease in its velocity from when it is thrown to just before it strikes the ground, so I imagine it is not perfectly elastic. But that is all the information given.

Here is the diagram. I am a newb and did not realize I could do this or I would have on my original post.

If you were to ask me, it looks like that horizontal line is just a reference to show that the 25.0 deg angle is with respect to the horizontal. In other words, the horizontal line is not part of the rock's path; it's just there for reference. I don't think there is a collision at all. To me it looks like the rock is initially launched, with its full initial speed, from a height of 4.60 meters.

So now apply your conservation of energy equation that you listed in the initial post. You've already calculated the kinetic energies (you might want to be more careful about your precision/rounding, btw.) You should have enough information now to calculate the potential energies.

 

[Honey Bee]

AH HAH! Man... Now I feel like a complete dumb dumb. THANK YOU so much... I believe you are correct. I shall work it out now.

I swear... physics is the easiest thing to over complicate.