A 5.0 kg block slides 10 m down arough 30 inclined plane. If the block wasinitia

Question

A 5.0 kg block slides 10 m down arough 30 inclined plane. If the block wasinitially
at rest, and a kinetic friction force of 10 N opposes the motion,calculate: (a) the work done by thefriction force.
(b) the work done by gravity.
(c) the work done by the normal force. (d) the final kinetic energyof the block. A 5.0 kg block slides 10 m down arough 30 inclined plane. If the block wasinitially
at rest, and a kinetic friction force of 10 N opposes the motion,calculate: (a) the work done by thefriction force.
(b) the work done by gravity.
(c) the work done by the normal force. (d) the final kinetic energyof the block.

Explanation / Answer

Work = Force X distance Force of friction = -10 N , opposes the motion Work = (-10N)(10m) W = -100 N*m Wgrav = -U We can find max height by using laws of trigonometry... 10 = length of hypotenuse, angle = 30 sin30 = h/10 10sin30 = h = 5 Wgrav = U1 - U2 U2 has height 0, at bottom of plane Wgrav = mgh - mg(0) Wgrav = (5kg)(9.8 m/s^2)(5m) Wgrav = 245 N*m Wnormal = Fdcos the angle between normal force and displacement = 0 Wnormal = Fdcos90 = 0. Wtotal = K Wtotal = Wgrav + W fric Wtotal = -100 + 245 Wtotal = 145 J Wtotal = K2 - K1, K1 = 0, object is at rest first 145J = KE final kinetic energy is 145J

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