CP CALC? A 5.00-kg crate is suspended from the end of a short vertical rope of negligible mass. An upward force F(t) is applied to the end of the rope, and the height of the crate above its initial position is given by y(t) = (2.80 m/s)t + (0.610 m/s3)t3. What is the magnitude of F when t = 4.00 s?

Solution 24E dy =v (y)=2.8 m/s+(1.83 m/s )t 3 2 dt dvy 3 3 dt =a (y)=3.66 m/s +(1.83 m/s )t , At t=4 s, dvy 2 dt =a (y)=14.64 m/s Newton's law in y direction F-mg=ma 2 F=49 N+(5 kg)(14.64 m/s ) =122 N The force is greater than the weight hence the crate accelerates upwards.