Q8E - University Physics with Modern Physics

Question

A loaded grocery cart is rolling across a parking lot in a strong wind. You apply a constant force $\overrightarrow{\mathbf{F}}\mathbf{=}\left(30N\right)\hat{\mathbf{i}}\mathbf{‐}\left(40N\right)\hat{\mathbf{j}}$ to the cart as it undergoes a displacement  $\overrightarrow{\mathbf{s}}\mathbf{=}\left(‐9.0m\right)\hat{\mathbf{i}}\mathbf{‐}\left(3.0m\right)\hat{\mathbf{j}}$. How much work does the force you apply do on the grocery cart?

Step-by-Step Solution

Answer

The total work applied by constant force to the grocery cart is -150 J.

Step 1: Identification of the given data

The given data is listed below as-

• The magnitude of the force acting on the cart is, $\overrightarrow{\mathrm{F}}=\left(30\mathrm{N}\right)\hat{\mathrm{i}}‐\left(40\mathrm{N}\right)\hat{\mathrm{j}}$  
• The displacement is, $\overrightarrow{s}=(-9.0 \mathrm{m})\hat{\mathrm{i}}‐(3.0 \mathrm{m})\hat{\mathrm{j}}$  

Step 2: Significance of the work done on a particle by a constant force

Work done on a particle by a constant force $\overrightarrow{\mathbf{F}}$ during a linear displacement $\overrightarrow{\mathbf{s}}$  is given by 

$\mathbf{W}\mathbf{=}\overrightarrow{\mathbf{F}}\mathbf{·}\overrightarrow{\mathbf{s}}$ ………….(1)

Step 3: Determination of work applied by constant force to the grocery cart

The work done can be obtained using equation (1), such that

For, $\overrightarrow{\mathrm{F}}=\left(30\mathrm{N}\right)\hat{\mathrm{i}}-\left(40\mathrm{N}\right)\hat{\mathrm{j}}$  and $\overrightarrow{\mathrm{s}}=\left(-9.0 \mathrm{m}\right)\hat{\mathrm{i}}-\left(3.0 \mathrm{m}\right)\hat{\mathrm{j}}$ 

The total work done will be

$$\begin{gathered} \mathrm{W}=\left({\mathrm{F}}_{\mathrm{x}}\hat{\mathrm{i}}+{\mathrm{F}}_{\mathrm{y}}\hat{\mathrm{j}}\right)·\left({\mathrm{s}}_{\mathrm{x}}\hat{\mathrm{i}}+{\mathrm{s}}_{\mathrm{y}}\hat{\mathrm{j}}\right) \\ ={\mathrm{F}}_{\mathrm{x}}{\mathrm{s}}_{\mathrm{x}}+{\mathrm{F}}_{\mathrm{y}}{\mathrm{s}}_{\mathrm{y}} \\ =\left(30 \mathrm{N}\times -9 \mathrm{m}\right)+\left(-40 \mathrm{N}\times -3 \mathrm{m}\right) \\ =-150\mathrm{N}·\mathrm{m} \\ \mathrm{W}=-150 \mathrm{J} \end{gathered}$$ 

 

The total work applied by constant force to the grocery cart is -150 J.