Recent content by Tinhorn

got it wasnt clear about the question but i got it

One last question if ΔKE = \frac{1}{2}mv^{2}_{x} + \frac{1}{2}mv^{2}_{y} why is ΔKE also -\frac{1}{2}mv^{2}_{x} + \frac{1}{2}mv^{2}_{y} should'nt it be -ΔKE

I get it. but i had always thought the x and y components of the force A is Ax= A*cosθ Ay= A*sinθ why does normal force have sin in its component and cos in its component

Can i say mgsinθ is the force moving the box down the slope since the x and y components of the force A is A_{x}= A*cosθ A_{y}= A*sinθ I can say mg sin θ = mg sinθcosθ + mgsin^{2} and since weight or gravity is also working on the box which is -mg on the y component. dot product...

so mg for weight and mg sin theta cos theta in the x mg cos ^2 theta in the y how does that make fr = mgsinθcosθ x+ mgsin^2y

mg in the -y direction that is why the sin is there mg sin theta cos theta in the y mg cos ^2 theta in the x

gravity has no x component it just has mg sin theta and normal force has mg sin theta cos theta + mg cos ^2 theta

Can you explain that to me again in greater detail please?

Do you know why the resultant force action on the box is this f_{r} = mgsinθcosθ x+ mgsin^{2}y I thought only mgsinθ and mgcosθ where did extra sin come in from

So in the rest frame, the incline isn't moving. So its KE = 0; in the moving frame, the incline is moving with v. so its KE = mv^{2}/2

isn't the force on the incline gravity

is the work done by both the negative force times the position

Lets say the velocity of the person is V_{fr} and the velocity of the box is V_{b} is the work on the incline m * V_{fr} - V_{b} and the work on the box the negative of that

Does it move backwards first and then from the top of the incline to the bottom.

it kinda looks likes a half circle First the box would be moving back and then slowing down and then acceleration