The force exerted by the motor on the cable is shown in the graph. Dtermine the velocity of the 105lb crate when t=2.5 sec

Q:-  The force exerted by the motor on the cable is shown in the graph. Dtermine the velocity of the 105lb crate when t=2.5 sec

Answer:- 

Given ,  

         Weight of the crate , W = 105 lb

Now, equation of the curve will be 

  F = 100t

Also, in Y- direction 

100t - 105 = 0

 t = 1.05 sec

So, time required to move the crate is 1.05 seconds

If the motion in Y- direction then, 

100t - 105 = ma

100t - 105 = (105/32.2) a

             a = 30.66t - 32.193

Now , we know that 

dV = adt

dV = (30.66t - 32.193)dt

after integration we get

V = 30.66t^2/2 - 32.193t

putting limits of t is 1.05 sec to 2.5 sec we get,

V = [30.66 (2.5)^2 /2 - 32.193(2.5)] - [30.66(1.05)^2 /2 - 32.193(1.05)]

V = [ 15.33 + 16.9013]

V = 32.2313 ft/sec

Hence, Dtermine the velocity of the 105lb crate when t=2.5 sec is 32.2313 ft/sec