Recent content by mybrohshi5

If this problem would have asked for the total work required to push the block up the ramp 15 m then would i have had to consider in the force of gravity like i did in my first post?

so if on my fbd i have Fg_x going down the ramp and F_pull going up the ramp i should have F_pull - Fg_x = ma then use 9.8m/s^2 for g?

then the forces in the y direction just equal zero right? cause N - Fg_y = 0 mgcos(theta) - mgcos(theta) = 0

I understand that the x force of gravity will be mg sin theta and the y will be mg cos theta I did draw a FBD. Mainly what i am confused about is adding or subtracting. So for the sum in the x is it F_pull + mgsin(theta) = ma or F_pull - mgsin(theta) = maand there is no friction :)

Homework Statement A box is being pulled up a ramp with a force F. The ramp has a 30 degree incline with the horizontal. write the sum of the forces in the x direction and y direction. Make the x and y coordinate system be parallel and perpendicular with the ramp. The Attempt at a...

x component of car B is actually -29228 so the total x component is -12353 so the total momentum of the system is P_total = sqrt ((46103^2) + (-12353^2)) P_total = 47730kgm/s :) Thank you

Homework Statement Two identical cars (m = 1350 kg) are traveling at the same speed (25 m/s). Car A is going 30 degrees East of North and car B is going 60 degrees West of North. What is the magnitude of the total momentum of the two-car system. Homework Equations p = mv The Attempt at a...

Oh so if W_friction = 1600J and it is done over a distance of 4 m then 1600 J = F_friction * 4m F_friction = 400 N Thank you :)

Oh yay so i wasnt too far off just a little mistake. Thank you Kuruman :)

Homework Statement A bird of mass 0.5kg is flying near to a building which is 35.0m tall. Take the direction AWAY FROM THE BUILDING to be the +x direction and take UP to be the +y direction. At some instant in time, the bird has an x velocity of +20m/s and a y velocity of -14m/s. At this same...

So if the boxes speed initially and finally are 0 then W = U + W_friction W = mgh 3200 = 800(2) + W_friction W_friction = 1600 J but that can't be right cause the answer is 400 J. I feel like i am being a complete idiot right now :(

ok so the boxes speed initially can be found using 1/2mv^2 = mgh v = 6.26 m/s the boxes final speed is just 0 correct? i don't feel like i am using the right initial and final kinetic's. This is what always seems to confuse me. I am not sure whether it will have 0 initial and a final...

My book states that the work-energy theorem is just W = delta K ? I think i am making this problem a lot harder than it actually is but i still can't figure out how to use what you have given me. I get it will have no potential energy at first but then it will have mgh potential energy when at...

Wow i just realized that i made that stupid mistake :) it is now edited for anyone looking at this in the future. Thanks Jay.

To be honest i am not sure why i did. i guess i was thinking that Work was force times the displacement of the x component of the ramp. The work-energy theorem is just W = delta K so W = K_f - K_i which is the same as 1/2mv_f^2 - 1/2mf_i^2 but i don't see how that can help me since i don't...