please answer questions 003, 004 and 005 questions 004 and 005 are related to 00
ID: 3163836 • Letter: P
Question
please answer questions 003, 004 and 005 questions 004 and 005 are related to 003Thanks
003 (part 1 of 3) 10.0 points A block is released from point A on a track ABCD as shown in the figure. Point A is higher than points B, C, and D. The track is frictionless except for a portion BC which has a coefficient of friction A. The block travels down the track and hits the spring with spring constant k. The acceleration of gravity is 9.8 m/s2 11 kg 3.5 m 3 m 191 kN/m BF C D Note: The distance h is not to scale. If the spring compresses 6 cm, find the coefficient of friction 004 (part 2 of 3) 10.0 points For this value of pu, to what height will the block rise when it returns to Point E Answer in units of m
Explanation / Answer
According to the given problem,
The velocity when it reaches the point B is given by
U2= 2gh = 19.6 x 4.3 = 84.28----------------1
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The work done in compressing the spring = 0.5 k x2;
This is equal to the kinetic energy 0.5 m V2
Hence V2 = k x2 / m= 191*103 * [0.06]2 / 11= 62.51 -----------2
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The energy is lost only in the portion BC
If U is the velocity at B and V at the point C
U2 – V2 = 21.77 -------[ From 1 and 2]
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From V2 – U2 = 2as = 2 a 3.5
a = - 21.77 / 7.0= -3.11 m/s2.
Hence =a/g = 0.317
Answer for a) = 0.317
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When the block moves from C to B, let W be its velocity at B.
W2 = V2 – 21.77 since it retunes through the same path.
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W2 = 62.51 - 21.77 = 40.74
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h is given by W2 = 2gh
h = 40.74 / 19.6 = 2.07 m.
Answer for b) = 2.07 m
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C)
At B. U2 = 2g 4.3 = 84.28
From B it travels a total distance of [3.5 + 3] = 6.5 and then stops.
The spring simply reverses the direction of velocity at C.
There is no loss of energy due to collision with the spring.
Hence U2 = - 2a [6.5]
84.28 =- 2 a [6.5]
a = -6.483 m/s^2
= =a/g = 0. 661
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