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Dmitry_Shevchenko [17]

3 years ago

11

5–111. A box having a weight of 8 lb is moving around in a circle of radius rA = 2 ft with a speed of (vA)1 = 5 ft>s while co

nnected to the end of a rope. If the rope is pulled inward with a constant speed of vr = 4 ft>s, determine the speed of the box at the instant rB = 1 ft. How much work is done after pulling in the rope from A to B? Neglect friction and the size of the box.

1 answer:

Elis [28]3 years ago

6 0

Answer:

a) vB = 10.77 ft/s

b) W = 11.30 lb*ft

Explanation:

a) W = 8 lb ⇒ m = W/g = 8 lb/32.2 ft/s² = 0.2484 slug

vA <em>lin</em> = 5 ft/s

rA = 2 ft

v <em>rad</em> = 4 ft/s

vB = ?

rB = 1 ft

W = ?

We can apply The law of conservation of angular momentum

L<em>in</em> = L<em>fin</em>

m*vA*rA = m*vB*rB ⇒ vB = vA*rA / rB

⇒ vB = (5 ft/s)*(2 ft) / (1 ft) = 10 ft/s (tangential speed)

then we get

vB = √(vB tang² + vB rad²) ⇒ vB = √((10 ft/s)² + (4 ft/s)²)

⇒ vB = 10.77 ft/s

b) W = ΔK = K<em>B</em> - K<em>A</em> = 0.5*m*vB² - 0.5*m*vA²

⇒ W = 0.5*m*(vB² - vA²) = 0.5*0.2484 slug*((10.77 ft/s)²-(5 ft/s)²)

⇒ W = 11.30 lb*ft

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