You are an industrial engineer with a shipping company. As part of the package-handling system, a small box with mass 1.60 kg is placed against a light spring that is com-pressed 0.280 m. The spring has force constant k = 45.0 Wm. The spring and box are released from rest, and the box travels along a horizontal surface for which the coefficient of kinetic friction with the box is 1.4 = 0.300. When the box has traveled 0.280 m and the spring has reached its equilibrium length, the box loses contact with the spring. (a) What is the speed of the box at the instant when it leaves the spring? (b) What is the maximum speed of the box during its motion?

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aristocles aristocles · Answer 1 · 2020-04-04T14:45:46+02:00

Answer:

Part a)

Velocity of the box is 0.74 m/s

Part b)

Maximum speed of the box is 0.93 m/s

Explanation:

Part a)

We can use the work energy theorem here to find the speed of the box

Now we can say that

work done by friction + work done by spring = change in kinetic energy

so we will have

[tex]-\mu mgx + \frac{1}{2}kx^2 = \frac{1}{2}mv^2[/tex]

now we have

[tex]-(0.300)(1.60)(9.81)(0.280) + \frac{1}{2}(45)(0.280^2) = \frac{1}{2}(1.60) v^2[/tex]

now we have

[tex]-1.32 + 1.76 = 0.8 v^2[/tex]

[tex]v = 0.74 m/s[/tex]

Part b)

When box will reach to the mean position then it will have maximum speed

so we have

[tex]\mu m g = kx[/tex]

[tex]0.300(1.60)(9.81) = 45 x[/tex]

[tex]x = 0.105 m[/tex]

now we can use work energy theorem

[tex]-\mu mg(x_1 - x_2) + \frac{1}{2}k(x_1^2 - x_2^2)= \frac{1}{2}mv^2[/tex]

[tex]-0.300(1.60)(9.81)(0.280 - 0.105) + \frac{1}{2}(45)(0.280^2 - 0.105^2) = \frac{1}{2}(1.6)v^2[/tex]

[tex]-0.825 + 1.516 = 0.8 v^2[/tex]

[tex]v = 0.93 m/s[/tex]