a 75 kg block is released from rest and then slides for adistance of 5.0m down a

Question

a 75 kg block is released from rest and then slides for adistance of 5.0m down a ramp that is inclined 30 degrees above thehorizontal. the 75 kg block strikes a 25 kg block that is restingon a horizontal surface at the bottom of the ramp. the block facesthat collide with each other are covered with Velcro causing thetwo blocks to stick together. Both the surface of the ramp and thehorizontal surface have a coefficient of kinetic friction of.15. a. what is the speed of the 75kg block at the bottom of theramp just prior to the collision with the 25 kg block? b. after the collision, how far do the two blocks travel onthe horizontal surface before coming to a stop a 75 kg block is released from rest and then slides for adistance of 5.0m down a ramp that is inclined 30 degrees above thehorizontal. the 75 kg block strikes a 25 kg block that is restingon a horizontal surface at the bottom of the ramp. the block facesthat collide with each other are covered with Velcro causing thetwo blocks to stick together. Both the surface of the ramp and thehorizontal surface have a coefficient of kinetic friction of.15. a. what is the speed of the 75kg block at the bottom of theramp just prior to the collision with the 25 kg block? b. after the collision, how far do the two blocks travel onthe horizontal surface before coming to a stop

Explanation / Answer

Given that the mass of block is m1 = 75 kg and m2 = 25kg   The height of ramp is h = 5.0 m    The coefficent of friction is = 0.15 ----------------------------------------------------------------   Then the length of ramp is S = h / sin= 5.0m / sin30o = 10m    Acceleration of the block on the ramp is a= g sin - *g cos                                                                  = 3.63 m/s2 Then the final velocity at the bottom of the ramp fromthe equation of motion                             v2 - u2 = 2aS                               v2- 0 = 2aS                                     v = 2aS                                         =8.52 m/s This is the speed of the 75 kg block at the bottom ofthe ramp just prior to the collision with the 25 kgBlock    Apply conservation of momentum before and afterthe collision             m1*v + 0 = (m1+m2)V                         V = m1*v / (m1+m2)                              =6.39 m/s This is the combained velcoity just aftercollision          From the workenergy theorem             workdone due to friction = change in kinetic energy                       -mg*x = (1/2)mVf2 -(1/2)mV2                         - mg*x = 0 - (1/2)mV2                                     x = V2 / 2*g                                     = 13.88 m    
Then the final velocity at the bottom of the ramp fromthe equation of motion                             v2 - u2 = 2aS                               v2- 0 = 2aS                                     v = 2aS                                         =8.52 m/s This is the speed of the 75 kg block at the bottom ofthe ramp just prior to the collision with the 25 kgBlock    Apply conservation of momentum before and afterthe collision             m1*v + 0 = (m1+m2)V                         V = m1*v / (m1+m2)                              =6.39 m/s This is the combained velcoity just aftercollision          From the workenergy theorem             workdone due to friction = change in kinetic energy                       -mg*x = (1/2)mVf2 -(1/2)mV2                         - mg*x = 0 - (1/2)mV2                                     x = V2 / 2*g                                     = 13.88 m    

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