Q59P - Fundamentals Of Physics

Question

A force ${\overrightarrow{\mathit{F}}}_{\mathit{a}}$ is applied to a bead as the bead is moved along a straight wire through displacement $\mathit{+}\mathit{5}\mathit{.}\mathit{0}\mathit{c}\mathit{m}$. The magnitude of  ${\overrightarrow{\mathit{F}}}_{\mathit{a}}$ is set at a certain value, but the $\mathit{\varphi }$ angle ${\overrightarrow{\mathit{F}}}_{\mathit{a}}$ between and the bead’s displacement can be chosen. Figure $\mathbf{7}\mathbf{-}\mathbf{45}$ gives the work $\mathit{W}$ done by  on the bead for a range of $\mathit{\varphi }$ values;  ${\mathit{W}}_{\mathit{0}}\mathit{=}\mathit{ }\mathit{25}\mathbf{ }\mathbf{J}$. How much work is done by  ${\overrightarrow{\mathbf{F}}}_{\mathbf{a}}\mathbf{ }$if $\mathit{\varphi }$ is (a) $\mathbf{64}\mathbf{°}$and (b) $\mathbf{147}\mathbf{°}$?

Step-by-Step Solution

Answer

1. Work done when $\varphi =64°$is $11 \mathrm{J}$.
2. Work done when $\varphi =147°$is $-21 \mathrm{J}$.

Step 1: Given information

It is given that, graph of work done with various values of, 

$$\begin{gathered} W_0=25\mathrm{J},\mathrm{when}\varphi =0, \\ \begin{array}{c}x=5.0 \mathrm{cm}\\ =0.05 \mathrm{m}\end{array}. \end{gathered}$$

Step 2: Determining the concept

The problem deals with the power and the work done. Work, energy, and power are the fundamental concepts of physics. Work is the displacement of an object when force is applied on it. By using the value of ${\mathit{W}}_{\mathit{0}}$ find the applied force ${\mathit{F}}_a$, which can be used to find the work done $\mathbf{64}\mathbf{°}$cat and $\mathbf{147}\mathbf{°}$.

 

Formula:

$W=Fd \cos \phi$

where,F is force, d is displacement and W is the work done

Step 3: (a) determining the work done when φ = 64 °

From the graph, 

$$\begin{gathered} W_0=25 \mathrm{J} \\ {W}_0=F_a.x \\ {F}_a=\frac{25}{0.05} \\ {F}_a=500 \mathrm{N} \end{gathered}$$

Hence, work done when $\varphi =64°$is $11 \mathrm{J}$.

Step 4: (b) determining the work done when ϕ = 147 ° .

Work done when $\varphi =64°$,

$$\begin{gathered} W=F_a\times \cos 64 \\ W=500\times 0.05\times 0.4383 \\ W=10.9575 \mathrm{J} \\ W\approx 11 \mathrm{J} \\ W=F_a\times \cos 147 \\ W=500\times 0.05\times (-0.8386) \\ W=-20.96 \mathrm{J} \\ W=-20.96 \mathrm{J} \\ W\approx -21 \mathrm{J} \end{gathered}$$

Hence, work done when $\varphi =147°$is$-21 \mathrm{J}$.