A block with mass m1 = 8.7 kg is on an incline with an angle = 38° with respect

Question

A block with mass m1 = 8.7 kg is on an incline with an angle = 38° with respect to the horizontal. For the first question there is no friction, but for the rest of this problem the coefficients of friction are: k = 0.34 and s = 0.374.

1) When there is no friction, what is the magnitude of the acceleration of the block?

2) Now with friction, what is the magnitude of the acceleration of the block after it begins to slide down the plane?

3) To keep the mass from accelerating, a spring is attached. What is the minimum spring constant of the spring to keep the block from sliding if it extends x = 0.16 m from its unstretched length.

4) Now a new block with mass m2 = 14.8 kg is attached to the first block. The new block is made of a different material and has a greater coefficient of static friction. What minimum value for the coefficient of static friction is needed between the new block and the plane to keep the system from accelerating?

Explanation / Answer

1) Apply, Net force acting on block, Fnet = m*g*sin(theta)

m*a = m*g*sin(38)

a = g*sin(theta)

= 9.8*sin(38)

= 6.03 m/s^2

2) Apply, Fnet = m*g*sin(theta) - mue_k*N

m*a = m*g*sin(38) - mue_k*m*g*cos(38)

a = g*sin(theta) - mue_k*g*cos(38)

= 9.8*sin(38) - 0.34*9.8*cos(38)

= 3.4 m/s^2

3) Apply, Fnet = 0

F_spring - m*g*sin(38) - mue_s*m*g*cos(38) = 0

F_spring = m*g*sin(38) + mue_s*m*g*cos(38)

k*x = m*g*sin(38) + mue_s*m*g*cos(38)

k = (m*g*sin(38) + mue_s*m*g*cos(38) )/x

= (8.7*9.8*sin(38) + 0.374*8.7*9.8*cos(38))/0.16

= 485 N/m

4) Tension in the string should be, T = k*x

= 485*0.16

= 77.6 N

so, now Apply net force on m2, Fnet2 = 0

m2*g*sin(38) + T - mue_s*m2*g*cos(38) = 0

mue_s = T/(m2*g*cos(38) + tan(38)

= 77.6/(14.8*9.8*cos(38) + tan(38)

= 0.674

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