A Block with mass 10kg is released from rest from the top of an incline plane th

Question

A Block with mass 10kg is released from rest from the top of an incline plane that has length = 15m. It makes an angle of theta = 53 degrees with the horizontal. The coefficient of friction between the block and plane is Mu = 0.40. 1. Draw a free body diagram 2. Derive an expression for the acceleration of the block down the plane. THEN calculate. aka don't plug in numbers until you derive the full equation. 3. Derive an expression for the speed of the block as it reaches the bottom of the plane. THEN calculate. Now, I don't really need help knowing what the answer is. What I absolutely need help on is how to derive these equations and find the components. Apparently for #2, it's F=ma. and you derive to get mg - fk = ma. f= MN, N = mgy = mgcostheta. it'll end up being mgsintheta - M(mgcostheta) = ma I need someone to go into great detail explaining how they got all those components i listed above. Not just an answer, I need a detailed explanation on finding components please. So it would be best if you move very slowly with the steps and write it out on a piece of paper. Thanks.

Explanation / Answer

See the weight of the block is acting downwards on incline which is mg ; Now Normal Reaction by incline = Weight component perpendicular to incline ; which comes out to be mg*cos 53 = N ; the other weight component mg*sin 53 is parallel to incline and hence is responsible for the motion of object ;

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