A box with mass m is dragged across a level floor with coefficient of kinetic friction ${\mathit{\mu }}_{\mathbf{k}}$by a rope that is pulled upward at an angle $\mathit{\theta }$above the horizontal with a force of magnitude F . (a) In terms of m , ${\mathit{\mu }}_{\mathbf{k}}$,$\mathit{\theta }$and g , obtain an expression for the magnitude of the force required to move the box with constant speed. (b) Knowing that you are studying physics, a CPR instructor asks you how much force it would take to slide a 90 - kg patient across a floor at constant speed by pulling on him at an angle of $\mathbf{25}\mathbf{°}$ above the horizontal. By dragging weights wrapped in an old pair of pants down the hall with a spring balance, you find that ${\mathit{\mu }}_{\mathbf{k}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{35}$. Use the result of part (a) to answer the instructor’s question

Mass = m

Coefficient of kinetic friction =${\mu }_k$

Angle =$\theta$

Force of magnitude = F

An object will accelerate in the direction of the net force, according to Newton's second law. This acceleration causes the object to slow down and eventually cease moving forward because the force of friction acts in the opposite direction to that of motion. Pushing with an amount of force equal to the amount of kinetic friction is necessary to keep it moving at a constant speed.

If the block slides with constant velocity then acceleration will be zero so that sum of all the forces will be zero and we break all forces in a horizontal and vertical component

In vertical direction-

$$\begin{gathered} F\sin \theta +N=mg \\ N=mg-F\sin \theta \end{gathered}$$ …(i)

In horizontal direction-

$F\cos \theta ={\mu }_{k}N$

From equation (i)

$$\begin{gathered} F\cos \theta ={\mu }_k\left(mg-F\sin \theta \right) \\ F\cos \theta +{\mu }_{k}F\sin \theta ={\mu }_{k}mg \\ F=\frac{{\mu }_{k}mg}{\cos \theta +{\mu }_k\sin \theta } \end{gathered}$$

Given

Mass = 90 kg

Angle$\theta =25°$

Coefficient of kinetic friction ${\mu }_k=0.35$

Using the above equation of force and put all these values-

$$\begin{gathered} F=\frac{0.35\times 90kg\times 9.8m/s^2}{\cos 25°+0.35\sin 25°} \\ =\frac{308.7}{1.054}N \\ =292.88N \end{gathered}$$

So that 292.88N force it would take to slide a 90 - kg patient across a floor at constant speed by pulling on him at an angle of $25°$above the horizontal.