Question

If the magnitude of the force she exerts on the suitcase is 25.0 N, and she does +1.12 X 10^3 J of work in moving the suitcase a distance of 51.0 m, at what angle above the horizontal (as shown in the figure above) is the force oriented with respect to the floor?degrees

Answer 1

Given data:

* The force acting on the suitcase is 25 N.

* The work done on the suitcase is,

$W=1.12\times10^3\text{ J}$

* The suitcase moved the distance is 51 m.

Solution:

The workdone in terms of force and displacement is,

$W=Fd\cos (\theta)$

where F is the force, d is the displacement,

$\theta\text{ is the angle between force and displacement,}$

Substituting the known values,

$\begin{gathered} 1.12\times10^3=25\times51\times\cos (\theta) \\ \cos (\theta)=\frac{1.12\times10^3}{25\times51} \\ \cos (\theta)=0.000878\times10^3 \\ \cos (\theta)=0.878 \\ \theta=28.6^{\circ} \end{gathered}$

Thus, the angle above the horizontal at which the force is oriented is 28.6 degree.