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*10^3\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*10^3=25*51*\cos (\theta) \\ \cos (\theta)=(1.12*10^3)/(25*51) \\ \cos (\theta)=0.000878*10^3 \\ \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.