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Block A (mass 40 kg) and block B (mass 80 kg) are connected by a string of negli

ID: 1283444 • Letter: B

Question

Block A (mass 40 kg) and block B (mass 80 kg) are connected by a string of negligible mass as shown in the figure. The pulley is frictionless and has a negligible mass. If the coefficient of kinetic friction between block A and the incline is ?k = 0.24 and the blocks are released from rest, determine the change in the kinetic energy of block A as it moves from C to D, a distance of 17 m up the incline.
______ J

Block A (mass 40 kg) and block B (mass 80 kg) are connected by a string of negligible mass as shown in the figure. The pulley is frictionless and has a negligible mass. If the coefficient of kinetic friction between block A and the incline is ?k = 0.24 and the blocks are released from rest, determine the change in the kinetic energy of block A as it moves from C to D, a distance of 17 m up the incline. ______ J

Explanation / Answer

Fy = n ? mg cos 37.0 = 0. Therefore,

n = mg cos 37.0 = 313.1 f = µn =0.240(313.1)= 75.14N

Wf = ?E or
(?75.14)(17) = ?UA + ?UB + ?K.EA + ?K.EB........eq1
where ?UA = mAg(hf ? hi) = (40)(9.80)(17sin 37.0)= 4×10^3J ...eq2

?UB =mBg(hf ? hi) = (80)(9.80)(?17)=?13.328×10^3J........eq3

Change in K.E. of A= 1/2 Ma (vf^2- vi^2)

Change in K.E.of B=1/2Mb (vf^2- vi^2)

Divide above two equation

Change in K.E of A / Change in K.E.of B= MA/MB=40/80=1/2

Change in K.E of A= Change in K.E of B/2.....eq 4

Substitute all the values in equation 1

-1277.38 = 4×10^3-13.328×10^3+?K.EA +2 ?K.EA

?K.EA= 2.683kJ