A block having a mass of 6 kg is projected with an initial velocity of 12 m/s, up an inclined plane. The plane is inclined at 45degre...?

Question

A block having a mass of 6 kg is projected with an initial velocity of 12 m/s, up an inclined plane. The plane is inclined at 45degrees and has a rough surface. The block travels 5 m up the plane and stops. Calculate:

a) the energy dissipated via the frictional forces; [224 J]

b) the magnitude of the average frictional force. [44.8 N]

Answer 1

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The weight m*g = 58.8N

The forces opposing motion are the weight and friction.

We can use conservation of energy to find friction losses.

KE initial = ½mV² = ½*6*12² = 432J (Mgh initial = 0)

Mgh final = 6*9.8*5*sin45 = 208J (KE final = 0)

432 - 208 = 224J <<<<< a)

224J = Ff*5m

Ff = 224J/5m = 44.8N <<<<< b) 

Answer 2

a. The energy dissipated by friction must equal the initial kinetic energy minus the final potential energy.  1/2 * 6 kg * 144 m2/s2 - 5 m * sin 45 * 9.81 m/s2 * 6 kg = E

E = 224 J

b. Fx = 224, F = 224/5 = 44.8 N