Two blocks are positioned on surfaces, each inclined at the same angle theta = B

Question

Two blocks are positioned on surfaces, each inclined at the same angle theta = B. 23.8 degree with respect to the horizontal. The blocks are connected by a rope which rests on a frictionless pulley at the top of the inclines as shown, so the blocks can slide together. The mass of the black block is C. 0.3 kg, and the coefficient of friction for both blocks and inclines is mu = D. 0.17. Assume gravity is g = 9.8 m/s^2. A) What is must be the mass of the white block if both blocks are to slide to the right at a constant velocity? ____.____ kg B) What is must be the mass of the white block if both blocks are to slide to the left at a constant velocity? ____.____kg C) What is must be the mass of the white block if both blocks are to slide to the right at an acceleration of 1.5m//s^2____._____kg D) What is must be the mass of white block if both blocks are to slide to the left at acceleration of 1.5m//s^2 ____.____kg E) Now, redo part D above, but now assuming no friction at all on either incline. So in the absence of friction, what must be the mass of the white block such that both blocks slide to the left at an acceleration of 1, 5m/s^2 ____.____kg

Explanation / Answer

a) tension in string for black block

mgsin23.8 = T + 0.17*mg*cos23.8

T = .3*9.8*sin23.8 - .17*.3*9.8*cos23.8

T = 0.73 N

for white block

T = 0.17Mw*gcos23.8 +Mw*gsin23.8

0.73 = Mw(0.17*9.8*cos23.8 + 9.8*sin23.8)

Mw = 0.13 kg

2)in this case

tension in string for black block

T = 0.17*mg*cos23.8 + mgsin23.8

= .17*.9.8*cos23.8 + .3*9.8*sin23.8

T = 2.7 N

for white block

Mwgsin23.8 = T + 0.17* Mwgcos23.8

Mw(9.8sin23.8 - 0.17*9.8*cos23.8) = 2.7

Mw = 1.11 kg

3)in this case

tension in string for black block

mgsin23.8 - T - 0.17mgcos23.8 = ma

T = .3*9.8*sin23.8 - 0.17*0.3*9.8*cos23.8 - .3*1.5

T = 0.27 N

for white block

.27 - (0.17Mw*9.8*cos23.8 +Mw*9.8*sin23.8) = Mw*1.5

0.27 = Mw(1.5 + 0.17*9.8*cos23.8 + 9.8*sin23.8)

Mw = 0.03 kg

4)in this case

tension in string for black block

T - (0.17*0.3*9.8*cos23.8 + .3*9.8sin23.8) = 0.3*1.5

T = 2.09 N

for white block

Mw*9.8*sin23.8 - (T + 0.17* Mw*9.8*cos23.8) = Mw*1.5

Mw(9.8*sin23.8 - .17*9.8*cos23.8 - 1.5) = 2.09

Mw = 2.24 kg

5) tension in string for black block

T - .3*9.8*sin23.8 = .3*1.5

T = 1.63 N

for white block

Mw*9.8*sin23.8 - T = Mw*1.5

Mw = 0.66 kg

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