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 connected 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.

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 lin = 5 ft/s

rA = 2 ft

v rad = 4 ft/s

vB = ?

rB = 1 ft

W = ?

We can apply The law of conservation of angular momentum

Lin = Lfin

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 = KB - KA = 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

any one of the statements could be an accurate comparison. but the statement that would define the correct choice and seal the deal is not even on the list:

"the visibilities of vector-x and vector-y in the drawing provided with the question are identical and nonexistent."