a block, m1, is resting on a rough horizontal surface. thecoefficient of static

Question

a block, m1, is resting on a rough horizontal surface. thecoefficient of static friction between the block and the horizontalsurface is .50. the coefficient of kinetic friction is .30. thisblock is connected to a block, m2, by a string over a frictionlesspulley. the block, m2, is hanging vertically. m1 has a mass of100kg. the system is released from rest. a) if the mass of m2 is 25kg, what is the static frictionforce on block m1? b)what is the minimum value of the mass of m2 that will juststart the system moving? c) what is the required value of the mass of m2 that will keepthe system moving with an acceleration of 2.0 m/s^2? PLEASE SHOW ALL WORK! Thank you a block, m1, is resting on a rough horizontal surface. thecoefficient of static friction between the block and the horizontalsurface is .50. the coefficient of kinetic friction is .30. thisblock is connected to a block, m2, by a string over a frictionlesspulley. the block, m2, is hanging vertically. m1 has a mass of100kg. the system is released from rest. a) if the mass of m2 is 25kg, what is the static frictionforce on block m1? b)what is the minimum value of the mass of m2 that will juststart the system moving? c) what is the required value of the mass of m2 that will keepthe system moving with an acceleration of 2.0 m/s^2? PLEASE SHOW ALL WORK! Thank you

Explanation / Answer

(25kg)(9.8m/s^2) = 245N

(b) The minimum value of m2 that will just start the system isobtained by sm1g =m2g Then sm1 =m2 Then m2 = (0.5)(100kg) =50kg
(c) For this we need m2g - T = m2a and T - km1g =m1a Then m2g - m1a - km1g = m2a then m2 = m1(a + kg)/(g - a)    = (100kg)[(2.0m/s^2) + (0.3)(9.8m/s^2)]/(9.8m/s^2 - 2.0 m/s^2)    = 63.33kg

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