5 lb boxes on an assembly line at 35 foot intervals. The boxes are moving at 12.

Question

5 lb boxes on an assembly line at 35 foot intervals. The boxes are moving at 12.5 ft/s when they are slowed by a motor. During the deceleration, the force applied to a box is F = t3/6 lb. After the boxes leave the decelerator at a much lower velocity, they slide down a smooth ramp to reach a new level. The height of the ramp is 18 in and the boxes are moving at a velocity of 10 ft/s at the bottom. (a) What was the speed of the box at the top of the ramp? (b) How long was the box slowed by the motor? Ignore friction effects on the ramp.

Explanation / Answer

a) At bottom of ramp v = 10 ft / sec

Using work energy theorem,

mgh = mv^2 /2 - mu^2 /2

32.17 x (18 / 12) = 10^2 /2 - u^2 /2

48.255 = 50 - u^2 /2

u = 1.87 ft/s

b) F = m(dv/dt) = -t^3 /6

6mdv = -t^3 dt

intergrating ,

6m ( vf - vi ) = -[ t^4 /4 ]

6 m ( 1.87 - 12.5) = - t^4 / 4

m = mass = 5lb

6 *5*4 *10.63 = t^4

t = 5.97 sec

if 5 lb is weight given not mass them m = 5/32.17

then, (5/32.17) x 6 x 4 x 10.63 = t^4

t = 2.51 sec

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