A constant force of magnitude 10 N makes an angle of $\mathbf{150}\mathbf{°}$(measured counterclockwise) with the positive x direction as it acts on a 2.0 kg object moving in an xy plane. How much work is done on the object by the force as the object moves from the origin to the point having position vector $\left(2.0m\right)\hat{\mathbf{i}}\mathbf{‐}\left(4.0m\right)\hat{\mathbf{j}}$?

It is given that,

The force acting on the object is$F=10\mathrm{N}$

Displacement of the object is from the origin to$2\hat{\mathrm{i}}-4\hat{\mathrm{j}}$

The problem deals with the work done which is the fundamental concept of physics. Work is the displacement of an object when force is applied on it. Find the force vector from its magnitude and angle made by it withthe+x axis. Using this force vector and displacement vector, find the work done using the corresponding formula.

Formula:

$W=\overrightarrow{F}.\overrightarrow{d}$

Where,

F is force, dis displacement and Wis the work done.

The force acting on the object is 10 N .

The components of the applied force are,

$$\begin{gathered} F_x=F\cos \theta \\ =10\cos \left(150\right) \\ =-8.66\mathrm{N} \\ {F}_y=F\sin \theta \\ =10\sin \left(150\right) \\ =5\mathrm{N} \end{gathered}$$

Hence, this force can be written as,

$\overrightarrow{F}=-8.66\hat{\mathrm{i}}+5\hat{\mathrm{j}}$

Displacement of the object is,

$$\begin{gathered} \overrightarrow{d}=\left(2-0\right)\hat{\mathrm{i}}+\left(-4-0\right)\hat{\mathrm{j}} \\ \overrightarrow{d}=2\hat{\mathrm{i}}-4\hat{\mathrm{j}} \end{gathered}$$

Thus, work done by the applied force is,

$$\begin{gathered} W=\overrightarrow{F}.\overrightarrow{d} \\ W=\left(-8.66\hat{\mathrm{i}}+5\hat{\mathrm{j}}\right)\left(2\hat{\mathrm{i}}-4\hat{\mathrm{j}}\right) \\ W=-37.32 \\ ~-37\mathrm{J} \end{gathered}$$

Thus, the work done on the object by the applied force is$-37\mathrm{J}$.