Problem 1 [Impulse and Momentum Method] A 1500 N block is in contact with a leve

Question

Problem 1
[Impulse and Momentum Method]
A 1500 N block is in contact with a level plane whose coefficientof kinetic friction is 0.10. If the block is acted upon by ahorizontal force of 250 N, what time will elapse before the blockreaches a velocity of 14.5 m/s, starting from rest? If the 250 Nforce is then removed, how much longer will the block continue tomove?

Problem 2
[Impact]
Direct central impact occurs between a 300 N body moving to theright at 3 m/s and a 150 N body moving to the left at 6 m/s. If thecoefficient of restitution is 0.6, what is the average impulsiveforce for a time of contact lasting 0.02 s?

Explanation / Answer

Mass (m) of the block = (1500N) / (9.8m/s2)                                  = 153.1 kg Coefficient of friction () = 0.10 Acceleration due to gravity (g) = 9.8m/s2 Force (F) acting on the block = 250N Initial speed (u) of the block = 0 Final speed (v) of the block = 14.5 m/s Net force acting on the block is          Fnet = F - f                 = F - mg                 = 250 N - (0.10)(1500N)                 = 100 N According to the Newton's second law of motion we have          Fnet = ma            a = (100N) / (153.1kg)              = 0.653 m/s2 We know the formula for the acceleration (a) of the block is            a = (v - u) / t            t = (v - u) / a              = (14.5m/s) / (0.653 m/s2)              = 22.2 s

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