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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...?
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]
2 Answers
- oldschoolLv 73 months agoFavorite Answer
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)
- JohnLv 73 months ago
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
