3) Masses m1 (2 Kg) and m2 (7 Kg) are connected by a 13 m long cord, hung over a

Question

3) Masses m1 (2 Kg) and m2 (7 Kg) are connected by a 13 m long cord, hung over a perfect pulley as shown. The coefficient of kinetic friction between m1 and the plane is 0.4. Mass m1 is released from pulley does not play a role. rest when m2 is located 6 m below the pulley. Note: The a) What is the normal force acting on the system? b) What is the friction force acting on the system? c) What is the Tension in the cord? d) What is the acceleration of the system? e) How long does it take for m1 to hit the pulley? f) What is the velocity of m1 as it hits the pulley?

Explanation / Answer

a) Normal force acting on the system is N = m1*g*cos(theta) = 2*9.81*cos(40) = 15 N

b) frictional force actiong on the system = mu_k*N = 15*0.4 = 6 N

c) Tension is T

T = m2*g = 7*9.8 = 68.6 N

d) accelaration of the system is a

Net force is Fnet = (T+m1*g*sin(40))-f

Fnet = (68.6+(2*9.8*sin(40))-6) = 75.2 N

m1*a = 75.2

a = 75.2/2 =37.6 m/s^2

e) initial velocity is Vo = 0 m/sec

s = 7 m

a = 37.6 m/s^2

then using

S = (Vo*t)+(0.5*a*t^2)

7 = (0*t)+(0.5*37.6*t^2)

t = 0.61sec

f) v = Vo +(a*t) = 0+(37.6*0.61) = 22.9 m/sec

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