# Newton's Laws: A 4.75 kg crate is suspended from the end of a short vertical rope of negligible mass.  A 4.75 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)= (2.80m/s)t + (0.61 m/s^3)t^3 What is the magnitude of the force F when 3.70s? answer in appropriate units

## Expert Answers

You need to find the net force acting on the crate, hence you need to use the equation of position and to differentiate it two times to find the acceleration such that:

`(dy)/(dt) = 2.80 + 1.83t^2`

`(d^2y)/(dt^2) = 3.66t `

You need to substitute `3.70`  for t in equation `(d^2y)/(dt^2) = 3.66t ` such that:

`(d^2y)/(dt^2) = 3.66*3.70 `

`(d^2y)/(dt^2) = a = 13.542 m*s^-1 `

Since you know the acceleration, you may find the net force multiplying the acceleration by the mass of crate such that:

`F=13.542*4.75 =gt F = 64.3245 `

You may find the magnitude of the force at `t=3.70 s`  adding the gravitatational force to the net force such that:

Gravity = `m*g =gt Gravity = 4.75*9.8 = 46.55 N `

Magnitude = `64.32 + 46.55 = 110.87 N `

Hence, evaluating the magnitude of the force at t=3.70 s yields 110.87 N.

Approved by eNotes Editorial Team

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