Recent content by brittkub1291

Okay, i got it. I just was doing the problem as if they gave me the height and the length, and i used pathagoreans theorem to find the hypotenuse so it was making my answers wrong.

Yeah, i just realized that. So the normal force is mgcos(x), the friction force should just be (.24)(mgcos(x)) right? It keeps coming out wrong. When the question say the hill is 500 m long would that be the hypotenuse of the triangle? I was thinking that it wouldn't be..

Okay, i tried finding the friction force by using F=ma but i don't know if acceleration is constant or not.

During a rockslide, a 340 kg rock slides from rest down a hillside that is 500 m long and 300 m high. The coefficient of kinetic friction between the rock and the hill surface is 0.24. (a) If the gravitational potential energy U of the rock-Earth system is set to zero at the bottom of the...

Okay, lol that was it. I think the main problem was i was thinking that this tree was like at an incline, like he was sliding down it like a slide. Yeah i deffinately overthought this one.

Okay, I'm thinking maybe i should just scratch using N2L, what if i just use the equation for gravitational energy, so Eg=(28kg)(9.8m/s)(6m) which gives me 1646.4. Then to find the change i would subtract the kinetic energy of the bear at the bottom of the slide to give me the change in...

Okay well i would have to use F=ma to find the frictional force right? I'm getting confused because i don't know the acceleration, should i just assume constant?

A 28 kg bear slides, from rest, 6 m down a lodgepole pine tree, moving with a speed of 5.9 m/s just before hitting the ground. (a) What change occurs in the gravitational potential energy of the bear-Earth system during the slide? (b) What is the kinetic energy of the bear just before...

Thanks for your help :]

Okay i think i figured it out. The force of gravity on the skier should be mgd and then the force of the rope on the skier should be the same. So the velocity doesn't matter, the work is the same either speed.

Ah, I'm not so good with the cutting and pasting apparently. Here's what it should be: c. At what rate is the force of the rope doing work on the skier when the rope moves with a speed of 2.0 m/s? And your right I'm not reading it correctly. So i figure now that since the work is 940 over...

Okay so for the first one i tried W=940Ncos(12)*8.8m, but it didnt work... And the second b should be a c lol sorry.

Okay so i think i start with P=Fvcos(x), so for the first one i plugged in the force, velocity and cosine of my angle and got the power. This didn't get me the answer, which is probably because I'm trying to find work. So i know that P=dw/dt, but i don't know the time so I'm kind of stuck.

A skier pulled by a tow rope up a frictionless ski slope that makes an angle of 12° with the horizontal. The rope moves parallel to the slope with a constant speed of 1.0 m/s. The force of the rope does 940 J of work on the skier as the skier moves a distance of 8.8 m up the incline. (a) If...

The only force acting on a 1.8 kg body as it moves along the positive x-axis has an x component Fx = - 6x N, where x is in meters. The velocity of the body at x = 3.0 m is 8.0 m/s. (a) What is the velocity of the body at x = 4.0 m? (b) At what positive value of x will the body have a...