In the figure below, the sliding block has a mass of 0.850 kg, the counterweight

Question

In the figure below, the sliding block has a mass of 0.850 kg, the counterweight has a mass of 0.420 kg, and the pulley is a uniform solid cylinder with a mass of 0.350 kg and an outer radius of 0.0340 m. The coefficient of kinetic friction between the block and the horizontal surface is 0.300. The pulley turns without friction on its axle. The light cord does not stretch and does not slip on the pulley. The block has a velocity of 0.940 m/s toward the pulley when it passes through a photo-gate.

(a) Use energy methods to predict the speed of the block after it has moved to a second photogate 0.700 m away.

(b) Find the angular speed of the pulley at the same moment.

Explanation / Answer

Mass of the block is  m_1 = 0.850kg mass of the ciounter weight is m _2 = 0.420kg mass of the solid cyllinder is  m_p = 0.350kg velocity of the block moving towards the pulley os v = 0.940m/s radiusof the pulley r_p = 0.0340m coefficient of kinetic friction between the block and horizontal surface is _k = 0.300 Moment of inertia is I = 1/2 m_p r_p^2       Angular velocity = v / r_p                                = 0.940m/s / 0.0340m                                = 27.64 rad /s Friction force retarding the sliding block is F_k = _k n                                                                           = 0.300 *0.850kg * 9.8 m/s^2                                                                          = 2.499 N    Potential energy is zero at the level of the counter weight where the sliding mass reaches the second photogate then from the work enrgy theorem W = (K.E_trans + K.E_rot + P.E ) _final - (K.E_trans + K.E_rot + P.E ) _initial - f_k * d = 1/2 (m_1+ m_2 )v_f^2 + 1/2 (1/2 * m_p r_p^2 )( vf^2 / r_p^2 ) + 0    - 1/2 (m_1 +m_2 )v_i^2 -1/2 (1/2 m_p r_p^2 ) ( v_i^2 / r_p^2 ) - m_2 *g*d solving we get v_f = v_i^2 + 2 ( m_2 - _k m_1 ) g * d / m_1 + m_2 + 1/2 m_p           = ( 0.940m/s)^2 + 2 ( 0.420kg - 0.300* 0.850kg ) *9.8 * 0.70m / 0.850 + 0.420 + 0.350/2           = 1.56m/s b ) Angular speed of the pulley _f = v_f / r_p                                                          = 1.56 m/s / 0.0340m                                                           = 46.0 rad /s

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