Question

A loaded grocery cart is rolling across a parking lot in a strong wind. You apply a constant force $\vec{F} =($ 33 $)\hat{i} - ($ 41 $)\hat{j}$ to the cart as it undergoes a displacement $\vec{s} = (-$ 9.4 $)\hat{i} - ($ 3.1 $)\hat{j}$ .

Part A

How much work does the force you apply do on the grocery cart?

Express your answer using two significant figures.

=		${\rm J}$

Answer 1

Concepts and reason

The concept required to solve the problem is the work done by a force.

Use the components of the vectors to calculate the dot product of the force and displacement vector that is work done.

Fundamentals

The cross product of two vectors is $\vec a = {a_x}\hat i + {a_y}\hat j + {a_z}\hat k$ and $\vec b = {b_x}\hat i + {b_y}\hat j + {b_z}\hat k$ is,

$\overrightarrow a \cdot \overrightarrow b = {a_x}{b_x} + {a_y}{b_y} + {a_z}{b_z}$

Here, ${a_x}$ is the x component of the vector $\overrightarrow a$ , ${a_y}$ is the y component of the vector $\overrightarrow a$ , ${a_z}$ is the z component of the vector $\overrightarrow a$ , ${b_x}$ is the x component of the vector $\overrightarrow b$ , ${b_y}$ is the y component of the vector $\overrightarrow b$ , ${b_z}$ is the z component of the vector $\overrightarrow b$ , $\hat i$ is the unit vector along x axis, $\hat j$ is the unit vector along y axis and $\hat k$ is the unit vector along z axis.

If a force displaces an object from its original position, then the work done on the object is,

$W = \overrightarrow F \cdot \overrightarrow s$

Here, $\overrightarrow F$ is the force acting on the object and $\overrightarrow s$ is the displacement of the object.

Part A

The force acting of the cart is,

$\overrightarrow F = 33\,{\rm{N}}\,\hat i - 41\,{\rm{N}}\,\hat j + 0\,{\rm{N}}\,\hat k$

The displacement of the cart is,

$\overrightarrow s = - 9.4\,{\rm{m}}\,\hat i - 3.1\,{\rm{m}}\,\hat j + 0\,{\rm{m}}\,\hat k$

The force in vector components is,

$\overrightarrow F = {F_x}\hat i + {F_y}\hat j + {F_z}\hat k$

Here, ${F_x}$ is the x component of the vector $\overrightarrow F$ , ${F_y}$ is the y component of the vector $\overrightarrow F$ , ${F_z}$ is the z component of the vector $\overrightarrow F$ .

The displacement in vector components is,

$\overrightarrow s = {s_x}\hat i + {s_y}\hat j + {s_z}\hat k$

Here, ${s_x}$ is the x component of the vector $\overrightarrow s$ , ${s_y}$ is the y component of the vector $\overrightarrow s$ , ${s_z}$ is the z component of the vector $\overrightarrow s$ .

Therefore the components of the force are,

$\begin{array}{c}\\{F_x} = 33\,{\rm{N}}\\\\{F_y} = - 41\,{\rm{N}}\\\\{F_z} = 0\,{\rm{N}}\,\hat k\\\end{array}$

The components of displacement are,

$\begin{array}{c}\\{s_x} = - 9.4\,{\rm{m}}\\\\{s_y} = - 3.1\,{\rm{m}}\\\\{s_z} = 0\,{\rm{m}}\\\end{array}$

The work done on the cart is,

$\begin{array}{c}\\W = \overrightarrow F \cdot \overrightarrow s \\\\\,\,\,\,\, = \left( {{F_x}\hat i + {F_y}\hat j + {F_z}\hat k} \right) \cdot \left( {{s_x}\hat i + {s_y}\hat j + {s_z}\hat k} \right)\\\\ = {F_x}{s_x} + {F_y}{s_y} + {F_z}{s_z}\\\end{array}$

Substitute $33\,{\rm{N}}$ for ${F_x}$ , $41\,{\rm{N}}$ for ${F_y}$ , $0\,{\rm{N}}$ for ${F_z}$ , $- 9.4\,{\rm{m}}$ for ${s_x}$ , $- 3.1\,{\rm{m}}$ for ${s_y}$ and $0\,{\rm{m}}$ for ${s_z}$ . The work done on the cart is,

$\begin{array}{c}\\W = \left( {33\,{\rm{N}}} \right)\left( { - 9.4\,{\rm{m}}} \right) + \left( { - 41\,{\rm{N}}} \right)\left( { - 3.4\,{\rm{m}}} \right) + \left( {0\,{\rm{N}}} \right)\left( {0\,{\rm{m}}} \right)\\\\ = - 183\,{\rm{J}}\\\end{array}$

The work done on the cart is $- 183\,{\rm{J}}$ .

Ans: Part A

The work done on the cart is $- 183\,{\rm{J}}$ .