A girl of mass mg = 40 kg is pulling a sled up a slippery slope. The coefficient

Question

A girl of mass mg = 40 kg is pulling a sled up a slippery slope. The coefficient of friction between the girl's boots and the slope is µs = 0.158; the friction between the sled and the slope is negligible. The girl can pull the sled up the slope with a = amax = 0.020 m/s2 before she begins to slip. Assume the rope connecting the girl to the sled is kept parallel to the slope at all times. The angle of the slope is ? = 8°. If the sled is being pulled up the slope with a=amax, what is the mass of the sled?

Link:
http://www.learning.physics.dal.ca/dalphysicslib/Graphics/Gtype10/sled.gif

Explanation / Answer

Given data Mass of the girl is, mg = 40 kg Let the mass of the sled be m, and the tension in the rope be T The coefficient of friction between the girl's boots and the slope is µs = 0.158 The acceleration of the sled is, a = 0.02 m/s2 Angle of inclination is, = 8o Solution: Using Newton's second law, the net force acting on the girl is, mga = T - mgg sin 8o - µsmgg cos 8o       T = mg ( a + g sin 8o + µsg cos 8o)          = 40 kg ( 0.02 m/s2 + ( 9.8 m/s2 )sin 8o + (0.158) ( 9.8 m/s2 )cos 8o )          = 116.68 N --------------------------------------------------------------------------------------------- The net force acting on the sled is,                          ma = T - mg sin 8o
   m ( a + g sin 8o ) = T                             m = 116.68 N / (0.02 m/s2 + ( 9.8 m/s2 )sin 8o )                                 = 84.31 kg
Thus, the mass of the sled is 84.31 kg

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