The question reads: A block is launched up africtionless ramp that makes an angl

Question

The question reads: A block is launched up africtionless ramp that makes an angle of 35 degrees to thehorizontal. If the block's initial speed is 2.2 m/s, how farup the ramp does it slide? I figured out the x and y components for the block so I couldrelate the two equations and solve for acceleration 'a', and then Iuse the equation V2=V02 +2axto find 'x', but my answer does not match the answer given in theback which is 43 cm. I keep coming up with 60 cm and I can'tfigure out what I'm doing wrong. The question reads: A block is launched up africtionless ramp that makes an angle of 35 degrees to thehorizontal. If the block's initial speed is 2.2 m/s, how farup the ramp does it slide? I figured out the x and y components for the block so I couldrelate the two equations and solve for acceleration 'a', and then Iuse the equation V2=V02 +2axto find 'x', but my answer does not match the answer given in theback which is 43 cm. I keep coming up with 60 cm and I can'tfigure out what I'm doing wrong.

Explanation / Answer

The first thing to realize is that the acceleration is actingopposite the direction of motion. If you take motion up theplane to be positive then the force of gravity acts downward andyour acceleration is: a=-g*sin() with =35 degrees For a constant force with no friction the expression to useis: v2-v02=2*a*(x-x0) You know the the block stops (they tell you this in theproblem by aking you how far it slides), so you know that the finalvelocity is zero. Your equation now becomes: -v02=2*a*(x-x0) Now rearrange and solve for (x-x0): (x-x0)=-v02/2*a=-v02/-2*g*sin(35)=.4305m=43.05cm Note that the negative signs cancel out. It always helps whenyou pick the right reference frame. Good Luck!!

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