Q88 P - University Physics with Modern Physics

Question

An object has several forces acting on it. One of these forces is $\overrightarrow{\mathbf{F}}\mathbf{=}\mathbf{\alpha xy}\hat{\mathbf{i}}$, a force in the x-direction whose magnitude depends on the position of the object, with . Calculate the work done on the object by this force for the following displacements of the object: (a) The object starts at the point (x = 0, y = 3.00 m) and moves parallel to the x-axis to the point (x = 2.00, y = 3.00 m). (b) The object starts at the point (x = 2.00, y = 3.00 m) and moves in the y-direction to the point, (x = 2.00, y = 3.00 m). (c) The object starts at the origin and moves on the line y = 1.5x to the point (x = 2.00, y = 3.00 m) .

Step-by-Step Solution

Answer

1. The work done is 15.0 J.
2. The work done is 0 J.
3. The work done is 10 J.

Step 1: Identify the given data

The force acting in the x-direction is $\overrightarrow{\mathrm{F}}=\mathrm{\alpha xy}\hat{\mathrm{i}}$ .

The value of $\mathrm{\alpha }=2.50\mathrm{N}/{\mathrm{m}}^2$.

The object starts at the point (x,y) = (0,3) and the object reaches the point (2,3).

Step 1: Concept/Significance of Work done

The expression of work done is given by,

$\mathbf{W}\mathbf{=}\mathbf{Fscos\theta }$…………………(1)

Step 3: Determine the work done on the object by force when the object starts at the point (x = 0, y = 3.00m ) and moves parallel to the x -axis to the point (x = 2.00, y = 3.00m) (a)

The work done can be calculated as follows.

$$\begin{gathered} \mathrm{W}={\int }_{{\mathrm{x}}_1}^{{\mathrm{x}}_2}\overrightarrow{\mathrm{F}}\mathrm{dx} \\ ={\int }_0^2\left(\mathrm{\alpha xy}\right)\mathrm{dx} \\ =(2.50 \mathrm{N}/{\mathrm{m}}^2)(3 \mathrm{m}){\int }_0^2\mathrm{xdx} \\ =\left(2.50 \mathrm{N}/{\mathrm{m}}^2\right)(3 \mathrm{m}){\left[\frac{{\mathrm{x}}^2}{2}\right]}_0^2 \end{gathered}$$

Simplify further,

$$\begin{gathered} \mathrm{W}=\left(2.50 \mathrm{N}/{\mathrm{m}}^2\right)(3 \mathrm{m})(2-0) \\ =15.0 \mathrm{J} \end{gathered}$$

Therefore, the required work done is .

Step 4: Determine the work done on the object by force when the object starts at the point (x = 2, y = 0 m) and moves parallel to the x-axis to the point (x = 2.00, y = 3.00 m) (b)

The object is moving in the perpendicular direction, so $\mathrm{\theta }=90°$ ., y

 

The work done can be calculated as follows,

 

$$\begin{gathered} \mathrm{W}=\mathrm{Fscos\theta } \\ =\mathrm{Fscos}\left(90°\right) \\ =0 \mathrm{J} \end{gathered}$$

 

Therefore, the required work done is 0 J .

Step 5: Determine the work done on the object by force when the object starts at the point (x = 0, y = 0 m) and moves along y = 1.5x to the point (x = 2.00, y = 3.00 m) (c)

The work done can be calculated as follows.

 

$$\begin{gathered} \mathrm{W}={\int }_{{\mathrm{x}}_1}^{{\mathrm{x}}_2}\overrightarrow{\mathrm{F}}\mathrm{dx} \\ ={\int }_0^2\left(\mathrm{\alpha xy}\right)\mathrm{dx} \\ =(2.50 \mathrm{N}/{\mathrm{m}}^2){\int }_0^2(1.5 \mathrm{x})\mathrm{xdx} \\ =\left(2.50 \mathrm{N}/{\mathrm{m}}^2\right)(1.5 ){\left[\frac{{\mathrm{x}}^3}{3}\right]}_0^2 \end{gathered}$$

 

Simplify further.

$$\begin{gathered} \mathrm{W}=\left(2.50 \mathrm{N}/{\mathrm{m}}^2\right)\left(1.5\right)\left(\frac{8}{3}-0\right) \\ =10.0 \mathrm{J} \end{gathered}$$

 

Therefore, the required work done is 10.0 J.