Q24E - University Physics with Modern Physics

Question

A 5.00-kg crate is suspended from the end of a short vertical rope of negligible mass. An upward force   is applied to the end of the rope, and the height of the crate above its initial position is given by $\mathit{y}\left(t\right)\mathbf{=}\left(2.80m/s\right)\mathit{t}\mathbf{+}\left(0.610m/s^3\right){\mathit{t}}^{\mathbf{3}}$ . What is the magnitude of  F when  $t=4.00s$ ?

Step-by-Step Solution

Answer

The magnitude of the force is 122 N.

Step 1: Identification of given data

The given data can be listed below as follows,

 

• The initial position of the crate, $y\left(t\right)=\left(2.80m/s\right)t+\left(0.610m/s^3\right)t^3$ .

 

• The weight of the crate, $m=5.00 kg$ .

Step 2: Significance of magnitude of the force

The magnitude of the force is the sum of all the forces that are acting on an object. It also tells the direction of forces which are applied to an object.

Step 3: Determination of magnitude of force.

The initial position of the crate is,

 

 $y\left(t\right)=\left(2.80m/s\right)t+\left(0.610m/s^3\right)t^3$

 

 Velocity is the derivative of position. Find the derivative of the above equation as:

 

 $$\begin{gathered} v\left(t\right)=\frac{d}{dt}\left[\left(2.80m/s\right)t+\left(0.610m/s^3\right)t^3\right] \\ v\left(t\right)=2.80m/s+1.83t^{2}m/s^3 \end{gathered}$$

 

Acceleration is the derivative of velocity. Find the derivative of the above equation as:

 

 $$\begin{gathered} a\left(t\right)=\frac{d}{dt}\left[2.80m/s+1.83t^{2}m/s^3\right] \\ a\left(t\right)3.66t m/s^3 \end{gathered}$$

 

The expression of the force is expressed as,

 

 $$\begin{gathered} F-mg=ma\left(t\right) \\ F-mg=m\left(3.66t\right) \end{gathered}$$

 

Here m  is the weight of the crate,  t is the time, and g is the acceleration due to gravity.

 

Substitute 5.00 kg  for m  and  4.00s for t  ,and$9.8m/s^2$for g  in the above equation, and we get,

 

 $$\begin{gathered} F-\left(5.00kg\right)\times \left(9.8m/s^2\right)=\left(5.00kg\right)\left(3.66\times \left(4.00s\right)\right) \\ F=122N \end{gathered}$$

 

Hence, the required force is 122 N.