please answer B through F and label the correct answer and shoe steps please. se
ID: 1789453 • Letter: P
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please answer B through F and label the correct answer and shoe steps please. second time posting need to see full steps and what answer corresponds to which question
50* 3m Block 1.02 m Block A is attached to a 3m long cord that is attached to a hook in the ceiling. The hook is also 3m from the ground as shown in the diagram. Block A is released from rest and accelerates downward in an arc At the lowest point in its arc, it hits block B and sticks to it, experiencing a perfectly inelastic collision. At this instant the cord breaks and the combined block A and block B system move to the right along a perfectly smooth (no friction) floor. A short distance later the combined object goes up a rough ramp, reaching a maximum distance up the ramp of 1.02 m. Block A has a mass of 8Kg. Block B has a mass of 2Kg A. B. C. (10PTS) How fast (speed) is Block A moving just before it collides with Block B? (10PTS) What is the tension in the cord just before Block A collides with Block B? (7PTS) How fast is the combined Block A and B object moving immediately after the collision? D. (10PTS)How much work is done by friction as the combined block A and B system moves up the ramp to its maximum position of 1.02m? (10PTS) What is the coefficient of kinetic friction on the ramp? E. Returning to the beginning of the problem and thinking only about the motion from when Block A was released to just before it collides with Block B, (3PTS) What would the speed of Block A have been just before the collision if a constant drag friction) force of 10N acted over the entire downward arc? F.Explanation / Answer
(A) Applying energy conservation,
PEi + KEi = PEf + KEf
- m g 3 cos50 + 0 = - m g (3) + m v^2 /2
v^2 /2 = 9.81 x 3 ( 1 - cos50)
v = 4.59 m/s .......Ans
(B) T - m g = m v^2 / L
T = (9) (9.81 + 4.59^2 / 3)
T = 151.4 N
(C) Applying momentum conservation,
8 x 4.59 + 0 = (8 + 2 )v
v = 3.67 m/s
(d) Applying w0rk- energy theorem,
Work done by gravity + work done by friction = change in KE
- 10 x 9.8 x 1.02 sin30 + W = 0 - 10 x 3.67^2 / 2
W = - 17.4 J
(e) W = - f . d
f = 17.4 / 1.02 = 17.02 N
u = 17.02 / (10 x 9.81 cos30) = 0.20
(f) d = 3 x (50/360) = 0.417 m
- 0.417 x 10 + (8 x 4.59^2 /2) = 8 v^2 / 2
v = 4.47 m/s
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