Work problem: refer to more details. (a trunk on a ramp...) (DOT PRODUCT)?

Question

A 35kg trunk is dragged 10m up a ramp inclined at an angle of 12deg to the horizontal by a force of 90N applied at an angle of 20deg to the ramp. At the top of the ramp, the trunk is dragged horizontally another 15m by the same force. Find the total work done. (What is the process of solving this? Using Dot Product)

But wat about Fg? I know how to the dot product and work and everything but I am gettinng stuck with fg...is work for fg: - Fa (25)..i knw dis is wrong dsnt match de answer, can u tell me what Work equals to here without putting any values in. W = Fpull (10) + ....

Answer 1

Work is done when a force if moved a distance. In differential terms:

dW = F dot dx , where F and dx are vectors.

In this case, the forces are piecewise constant, so that you get the total work by adding that done going up the ramp plus that done on the horizontal.

I see no mention of friction, so that makes it easier.

Sketch a free body diagram with all the forces acting on the block. Add up all vector forces to get the net force. Get the component of this along the ramp and multiply by the ramp length to get work. Note that this is the same as taking the dot produce of the ramp length and the net force vector. That's where your dot product comes in.

Put it together, repost, then I'll check what you've done.

Answer 2

dot product is a good idea.
♠ slope: Let vector r is the path or r=x·i+y·j =|r|(cos(θ)·i+sin(θ)·j), where |r|=10m,
θ=12°; you are pulling the trunk m=35kg holding the rope end behind you; thus F=|F|(cos(σ)·i+sin(σ)·j), where |F|=90N, σ=12+20 =32°;
slope work Ws = (F·r) = |F|*|r|*( |(cos(θ)·cos(σ) + sin(θ)· sin(σ)) = ???
♣ horiz: r=15·i; F=90(cos(20°)·i +sin(20°)·j);
Wh=15*90* cos(20°) = ???
■ total W=Ws+Wh = ???; given m=35kg is baffle you! And what says the book back?