## two block system

in a two block system, block 1 lies on a horizontal surface, block two is attached to block one over a pulley and hanging down. Both blocks are 1kg in mass. there is friction between the block and the board. Block one has just been pushed to the left, released, and it is now slowing down. The coeeficient of friction between block one and the board is 0.15

a.) calculate normal force between sliding block and board, and then the magnitude of the frictional force

b.) write down newtons second law as it applies to each block

c.) determine the acceleration of the system and the tension in the string

• Anonymous commented

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• a)

Normal force = m1*g = 1*9.81 = 9.81 N

Friction force F = fric coeff*Normal force = 0.15*9.81 = 1.47 N

b) Let tension in string be T.

Block 1: T - F = m1*a

Block 2: m2*g - T = m2*a

c) Adding, m2*g - F = (m1 + m2)*a

1*9.81 - 1.47 = (1 + 1)*a

a = 4.17 m/s^2

T = F + m1*a = 1.47 + 1*4.17 = 5.64 N

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