A block of mass m 1 = 2.00 kg is placed in front of a block of mass m 2 = 5.75 k

Question

A block of mass m1 = 2.00 kg is placed in front of a block of mass m2 = 5.75 kg, as shown in the figure. The coefficient of static friction between m1 and m2 is 0.63, and there is negligible friction between the larger block and the tabletop. (Take the +x-direction to be to the right.)

(a) What forces are acting on m1? (Select all that apply.)

(b) What is the minimum external force F that can be applied to m2 so that m1 does not fall?
2_____________

(c) What is the minimum contact force between m1 and m2?
3_____________

(d) What is the net force acting on m2 when the force found in part (b) is applied?
_____________

Please be very clear with the answers..

Explanation / Answer

m1 = 2.00 kg, m2 = 5.75 kg, mu = 0.63, (a) What forces are acting on m1? frictional forces gravitational forces a contact force (b) What is the minimum external force F that can be applied to m2 so that m1 does not fall? F = (m1 + m2)a, contact force N = m1*a = m1*F/(m1 + m2) maximum static friction force f = mu*N = m1*g mu*m1*F/(m1 + m2) = m1*g so F = (m1 + m2)g/mu = 121 N (c) What is the minimum contact force between m1 and m2? N = m1*F/(m1 + m2) = 31.1 N (d) What is the net force acting on m2 when the force found in part (b) is applied? m2*a = m2*N/m1 = 89.4 N

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