a block of 4.4kg is held at the top of a smooth incline. the incline has height

Question

a block of 4.4kg is held at the top of a smooth incline. the incline has height of 2.1 m and is at an angle of 19 degrees above the horizontal. it is allowed to slide down the incline. A. what is the kinetic energy, K, the potential energy, U, and the total mechanical energy, E, of the block at the top of the incline? B. what are K, U, and E of the block when it reaches the bottom of the incline? C. if the incline were not smooth, the velocity of the block at the bottom of the incline would be 4.1 m/s. How much work was done by friction on the block?

Explanation / Answer

(a) K = 0 U = mgh = 4.4*9.8*2.1 = 90.55 J E = U + K = 90.55 J (b) As the block slides down the incline all the potential energy gets converted to K.E so, K = 90.55 J U = 0 E = U + K = 90.55 J (c) K.E at bottom = .5*m*v^2 = 36.98 J = K , Loss of P.E = mgh = 90.55 J = U so, W.D by friction = U-K = 53.57 J

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