Friction and Newton’s Laws PRACTICE A block of mass m1 = 300 g is sitting on a t

Question

Friction and Newton’s Laws PRACTICE A block of mass m1 = 300 g is sitting on a table with coefficient of kinetic friction k = 0.5. Attached to this block is a massless string which passes through a frictionless pulley, mounted on the edge of the table. Attached to the other end of the string is a second block of mass m2 = 2 kg. Suppose m2 falls under the influence of gravity, pulling m1 across the table. But suppose also that m2 experiences an upward force due to air resistance, Fair = 10 N. If the second block begins at rest at a height h0 = 2 m, calculate the time it takes to hit the ground.

Explanation / Answer

Normal force on m1 = 0.3*9.8 = 2.94 N

so friction = normal * k = 2.94*0.5 = 1.47 N

Weight of m2 = 2*9.8 = 19.6 N

Air resistance = 10 N

So effective downward force = 19.6 - 10 - 1.47 = 8.13 N

effective mass = 2 + 0.3 = 2.3 Kg

So acceleration of system = 8.13 / 2.3 = 3.535 m/s2

Height =2 m

initial velocity = 0

So applying, S = ut + 0.5at2

=> 2 = 0 + 0.5 * 3.534 *t2

=> t = 1.064 s

So the time it takes to hit the ground = 1.604 s

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