VA 1101 Sp 17: Lecture Quiz x C GOAL Use Conservation x C www.webassign. t/web/S
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VA 1101 Sp 17: Lecture Quiz x C GOAL Use Conservation x C www.webassign. t/web/Student/Assignment-Responses/submit?dep 15436575 Use the worked example above to help you solve this problem. A block with mass of 5.79 kg is attached to a horizontal spring with spring constant k 3.13 x 102 N/m, as shown in the figure. The surface the block rests upon is frictionless. The block is pulled out to xE 0.0550 m and released. (a) Find the speed of the block at the equilibrium point. 4043 (b) Find the speed when x 0.031 m. Your response is within 10% of the correct value. This may be due to roundoff error, or you could have a mistake in your calculation. Carry out all intermediate results to at least four-digit accuracy to minimize roundoff error. m/s (c) Repeat part (a) if friction acts on the block, with coefficient put 0.120. Your response differs from the correct answer by more than 10%. Double check your calculations m/s EXERCISE HINTS GETTING STARTED I M STUCK Use the values from PRACTICE IT to help you work this exercise. Suppose the spring system in the last example starts at x 0 and the attached object is given a kick to the right, so it has an initial speed of 0.54 m/s. (a) What distance from the origin does the object travel before coming to rest, assuming the surface is frictionless? What is the total energy of the system just after the block is given the kick? What is the total energy when the block is at rest? Do any forces do work on the system during this process m (b) How does the answer change if the coefficient of kinetic friction is u 0,120 Use the quadratic formula.) P A 3/2/2017Explanation / Answer
Using law of conservation of energy
Total energy at the extreme point = Total energy at the Equilibrium point
0.5*k*A^2 = 0.5*m*V^2
3.13*10^2*0.055^2 = 5.79*V^2
V = 0.404 m/sec
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b) Using law of conservation of energy
Total energy at the extreme point = Total energy at x = 0.031 m
0.5*k*A^2 = (0.5*m*v^2) +(0.5*k*x^2)
3.13*10^2*0.055^2 = (5.79*v^2)+(0.5*3.13*10^2*0.031^2)
v = 0.37 m/sec
C) Work done by the net force = change in kinetioc energy
Work done by the (spring Force + frictional force) = 0.5*m*(V^2-Vo^2)
initial speed is Vo = 0 m/sec
then
0.5*k*A^2 - (mu_k*m*g*S) = 0.5*m*V^2
(0.5*(3.13*10^2*0.055^2)) -(0.12*5.79*9.8*0.055) = 0.5*5.79*V^2
0.0989 = 0.5*5.79*V^2
V = 0.184 m/sec
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Exercise :
a) Using law of conservation of energy
Energy at the x = 0 = Energy at the extreme position
0.5*m*V^2 = 0.5*k*A^2
0.5*5.79*0.54^2 = 0.5*3.13*10^2*A^2
A = 0.0734 m
b) Work done by the net force = change in kinetic energy
-0.5*k*A^2 - (mu_k*m*g*S) = 0.5*m*(V^2-Vo^2)
(-0.5*3.13*10^2*0.0734^2)-(0.12*5.79*9.8*S) = 0.5*5.79*(0^2-0.54^2)
S = 0.000152 m
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