Two blocks of m1 and m2 are connected by a string and placed on a flat surface.

Question

Two blocks of m1 and m2 are connected by a string and placed on a flat surface. They are initially at rest. The friction coefficient = 0.2. A force F is exerted on second block. m1= 1 kg F is either 50 N, 20 N, or 4 N m2 = 1.5 kg t1= 4 s t2 = 7 s a) What is the velocity of the blocks at time t1? b) What is the tension of the string T? c) What is the momenta and kinetic energies of the blocks at time t2? Two blocks of m1 and m2 are connected by a string and placed on a flat surface. They are initially at rest. The friction coefficient = 0.2. A force F is exerted on second block. m1= 1 kg F is either 50 N, 20 N, or 4 N m2 = 1.5 kg t1= 4 s t2 = 7 s a) What is the velocity of the blocks at time t1? b) What is the tension of the string T? c) What is the momenta and kinetic energies of the blocks at time t2?

Explanation / Answer

Considering both the masses as our system, Net force = F - frictional force =F - 0.2(1+1.5)*9.8 =F - 4.9 N So when F =50 N Net force = 45.1 N Net acceleration = Force / mass = 45.1 / 2.5 = 18.04 m/s2 So velocity after 4 second = u +at = 0 + 18.04*4=72.16 m/s b)Tension in string = m1*acceleration = 1*18.04 = 18.04 N c)Velocity after 7 second = u +at = 0+18.04*7= 126.28 m/s So momentum = mass*velocity Kinetic energy = mass*velocity2/2 Put mass =2.5 kg to solve When force is 20 N ,use the same formula and method as for F=50 N When force = 4 N Here applied force < maximum static force So frictional force = 4 N and the bodies will not move , so velocity after time (t=4 sec) = 0 Tension = 4 - 0.2*9.8*1.5 = 1.06 N Also as blocks are not moving , so momentum and kinetic energy will also be zero

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