91Ó°ÊÓ

Mass = m

Coefficient of kinetic friction =${\mathrm{μ}}_k$

Angle =$\mathrm{θ}$

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-

$$\begin{gathered} F\sin \mathrm{θ}+N=mg \\ N=mg-F\sin \mathrm{θ} \end{gathered}$$ …(¾±)

In horizontal direction-

$F\cos \mathrm{θ}={\mathrm{μ}}_{k}N$

From equation (i)

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

Given

Mass = 90 kg

Angle$\mathrm{θ}=25°$

Coefficient of kinetic friction ${\mathrm{μ}}_k=0.35$

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

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

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.