What you're saying is that you would set its KE equal to its PE?
Did you actually work out the problem? Because I couldn't get an answer by doing the algebra. I did it several times.
Okay, but I was told to use the conservation of energy first (the kinetic energy of the ball as soon as it is released is equal to its initial potential energy...so you can find it's initial velocity). I set its initial kinetic energy equal to the final kinetic energy of the two balls combined...
Here's the problem:
"Two steel balls are suspended on (massless) wires so that their centers align. One ball, with mass 2.30 kg, is pulled up and to the side so that it is 0.0110 m above its original position. Then it is released and strikes the other ball in an elastic collision. If the...
By using the pythagorean theorem and the law of sines, I found the normal force to be approx. 5.17 N, which I multiplied by 11.0 m to get 56.92 J. Correct? Then I inserted that in the equation:
KE_i + PE_i + (normal force x hypotenuse) = KE_f
But I got the wrong answer. What am I doing wrong?
I'm having trouble with the following problem:
"There is a car ready to go down a hill. The height of the hill is 9.0 m, and the length of the slope is 11.0 m (hypotenuse). The frictional force opposing the car is 125 N, and the car must be going 12.5 m/s when it reaches the bottom. What...
Got it.
The forces exerted on the person are the tension of the rope in the upward direction, the force of the seat in the upward direction, and its weight (mg) in the downward direction. So,
T + F_seat + (-mg) = ma
348 N + F_seat - 612 N = (62.45 kg)(1.06 m/s^2)
F_seat = 330 N
Thanks!
Here's the question:
A 612 N person is sitting in a 16.0 N seat with one end of a massless rope attached to the seat. The rope runs over an ideal pulley, and the other end of it is in the person's hand. The person pulls the loose end of the rope with a force of 348 N.
The problems asks to...
Okay thanks, that makes sense. I was just getting confused about why T is multiplied by sin\theta. I was solving it a different way and didn't see that. Thanks :smile:
So I can solve for the combined tension and then divide that answer by 2 (since they're symmetrical), or should I divide earlier in the problem? (I have another problem that involves a non-symmetrical arrangement, so I'm trying to figure out when to make the distinction.)
I'm working on the following question:
"A painting of mass 3.20 kg hangs on a wall. Two thin pieces of wire, each 0.250 m long, connect the painting's center to two hooks in the wall. The hooks are at the same height and are 0.330 m apart. When the painting hangs straight on the wall, how...
This one should be quick and easy, but I just can't seem to get it:
"A college rower can easily push a small car along a flat road, but she cannot lift the car in the air. Since the mass of the car is constant, how can you explain this discrepancy?"
Edit: I'm thinking that friction can be...