Question

A horse pulls a cart with a force of $\mathbf{40}\mathbf{}\mathbf{lb}$at an angle of${\mathbf{30}}^{\mathbf{°}}$above the horizontal and moves along at a speed of$\mathbf{6}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{mi}\mathbf{/}\mathbf{h}$. (a) How much work does the force do in$\mathbf{10}\mathbf{}\mathbf{}\mathbf{min}$? (b) What is the average power (in horsepower) of the force?

Short answer

• Work done by the force in 10 min is$2.48\times {10}^5\mathrm{J}$
• Average power (in horsepower) of the force is$0.54\mathrm{hp}$

Step-by-step solution

Given information

It is given that,

The force is$F=40\mathrm{lb}=177.9289\mathrm{N}$

The time interval is $∆t=10\min =600\mathrm{s}$

The angle between $\stackrel{\rightharpoonup }{F}$and $\stackrel{\rightharpoonup }{d}$is$\varnothing ={30}^{°}$

Speed of cart is $v=6\mathrm{m}/\mathrm{hr}=2.68224\mathrm{m}/\mathrm{s}$

Determining the concept

First, calculate the value of displacement $\mathit{d}$. By using the value of $\mathit{d}$,the given value of force ${\mathit{F}}_{\mathbf{y}}$and the angle $\mathit{\varnothing }$between localid="1657181022004" $\stackrel{\mathbf{\rightharpoonup }}{\mathit{F}}$andlocalid="1657181079516" $\stackrel{\mathit{\rightharpoonup }}{\mathit{d}}$, calculate the value of work done $\mathit{W}$by the force$\mathit{F}$. By using this value of $\mathit{W}$, find the value of the average power ${\mathit{P}}_{\mathbf{a}\mathbf{v}\mathbf{g}}$.

Formulae:

The displacement is

$d=v∆t$

The work done by the force$\stackrel{\rightharpoonup }{F}$is

$$\begin{gathered} W=\stackrel{\rightharpoonup }{F.}\stackrel{\rightharpoonup }{d} \\ =Fd\cos (\mathit{\varnothing }\mathit{)} \end{gathered}$$

The average power is

$P_{avg}=\frac{w}{∆t}$

Where,$\stackrel{\rightharpoonup }{F}$is force vector, d is displacement, Wis the work done, vis velocity and tis time.

(a) Determining the work done by the force in 10 min

Now, if $∆t=10min=600s$then displacement $d$is$d=v∆t$

$d=2.68224\times 600$

$d=1609.344\mathrm{m}$

The work done by the force $\stackrel{\rightharpoonup }{F}$is,

$$\begin{gathered} W=\stackrel{\rightharpoonup }{F}.\stackrel{\rightharpoonup }{d} \\ =Fdcos(\mathit{\varnothing }\mathit{)} \\ W\mathit{=}177.9289\times 1609.344\times \cos \left(30\right) \\ W\mathit{=}247985.3418\mathrm{J} \\ \mathit{}\mathit{}\mathit{}\mathit{=}2.48\times {10}^5\mathrm{J} \end{gathered}$$

Hence, work done by the force in 10 min is$W=247981.64\mathrm{J}=2.48\times {10}^5\mathrm{J}$

(b) Determining the average power (in horsepower) of the force

The average power is given by,

$$\begin{gathered} P_{avg}=\frac{W}{∆t} \\ {P}_{avg}=\frac{247985.3418}{600} \\ {P}_{avg}=413.3089\mathrm{W} \\ {P}_{avg}=\frac{413.3027}{746}\mathrm{hp} \\ {P}_{avg}=0.55\mathrm{hp} \end{gathered}$$

Hence, average power (in horsepower) of the force is $P_{avg}=0.54\mathrm{hp}$.