6. 0.84/1.25 points| Previous Answers SerPSET9 5.P.050. My Not In the Atwood mac

Question

6. 0.84/1.25 points| Previous Answers SerPSET9 5.P.050. My Not In the Atwood machine shown below, m2.00 kg and m6.30 kg. The masses of the pulley and string are negligible by comparison. The pulley turns without friction and the string does not stretch. The lighter object is released with a sharp push that sets it into motion at vi = 2.35 m/s downward t1 m2 (a) How far will mi descend below its initial level? .5414 m (b) Find the velocity of mi after 1.80 s magnitude directionupward 1.417 Your response differs from the correct answer by more than 10%. Double check your caculetions. m/s

Explanation / Answer

Suppose, 'a' is the acceleration of the system.

So, we have the following expressions -

a = (T - 2g)/2 = (6.3g - T)/6.3

=> 6.3*(T - 2g) = 2*(6.3g - T)

=> T*(6.3 + 2) = 12.6g + 12.6g

=> T = 25.2g / 8.3 = 29.78 N

So, a = (29.78 - 2*9.81) / 2 = 5.08 m/s^2

(a) Distance descended by mass m1 -

y = Vi²/(2a) = 2.35^2 / (2*5.08) = 0.543 m

(b) Velocity of m1 after 1.80 s -

V = -Vi + a*t = -2.35 + 5.08*1.80 = + 6.80 m/s

Positive sign indicates that the velocity is in the upward direction.

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