Q50E - University Physics with Modern Physics

Question

A crate on a motorized cart starts from rest and moves with a constant eastward acceleration of $\mathit{a}\mathbf{=}\mathbf{2}\mathbf{.}\mathbf{80}\mathbf{ }{\mathbf{\text{m/s}}}^{\mathbf{\text{2}}}$ . A worker assists the cart by pushing on the crate with a force that is eastward and has magnitude that depends on time according to $\mathit{F}\mathbf{\left(}\mathit{t}\mathbf{\right)}\mathbf{=}\left(5.40 \text{N/s}\right)\mathit{t}$ . What is the instantaneous power supplied by this force at $\mathit{t}\mathbf{=}\mathbf{5}\mathbf{.}\mathbf{00}\mathbf{ }\mathbf{\text{s}}$?

Step-by-Step Solution

Answer

$378 \text{W}$

Step 1: Identification of given data

Acceleration of car is $a=2.80 {\text{m/s}}^{\text{2}}$

Time dependent force is $F\left(t\right)=\left(5.40 \text{N/s}\right)t$

Step 2: Significance of power

Power is defined as work done per unit time. It is given as

$\mathit{P}\mathbf{=}\frac{\mathit{W}}{\mathit{t}}\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{\text{or}}\mathbf{ }\mathbf{ }\mathit{F}\mathit{v}$                                                                                                                              ...(i)

Where, $W$ is the work done, $t$ is the time , $F$ is the force and $v$ is the velocity

Step 3: Determining the instantaneous power supplied by this force at t = 5 . 00   s

The final velocity of the crate is expressed as follows

$v = u + at$

Where $u$ is the initial velocity

In the question, it is given that the cart start from rest so that the initial velocity $u = 0$

So,

$v = at$                                                                                                                                       ...(ii)

By using equation (i) and equation (ii)

$\begin{array}{rcl}P& =& Fat\\ & =& \left(5.40 \text{N/s}\right)t\times at\\ & =& a\times \left(5.40 \text{N/s}\right)t^2\end{array}$

Substitute the values in above equation

$\begin{array}{rcl}P& =& 2.80 {\text{m/s}}^{\text{2}}\times 5.40 \text{N/s}\times {\left(\text{5.00} \text{s}\right)}^2\\ & =& 378 \text{W}\end{array}$

Hence the instantaneous power supplied by this force at $t=5.00 \text{s}$ is $378 \text{W}$