A 14-lb crate is pulled up a frictionless 40° ramp with an initial velocity of v1=0.4 ft/s. It is pulled 0.3 ft from location #1 to #2 by a pulley system and a P=36-lb force pulling down on pulley A. Also, a spring (k=69 lb/ft) is attached
to the crate which is un-stretched at location #2. Assume the spring, cable, and pulleys are massless. Find the final velocity of the crate in ft/sat location #2.

3.25 ft/s

Explanation:

The crate is of =14-lb=m₁

The angle of inclination is = 40°=Ф

The initial velocity = 0.4 ft/s= v₁

Distance the crate will move is= 0.3 ft =d

The load pulling downwards is = 36 lb= m₂

Acceleration of the pulley, a= m₂g - m₁gsinФ / m₁+m₂ where g= 32.17 ft/s^2

a= 36*32.17 - 14*32.17*sin 40° / 14+36

a=17.37 ft/s^2

Apply the formula for final velocity

V₂²=V₁²+2ad

V₂²=0.4²+ 2*17.37*0.3

V₂²=10.582

V₂ =√10.582 = 3.25 ft/s

answer:

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