You pull a short refrigerator with a constant force $\stackrel{\mathbf{\rightharpoonup }}{\mathbf{F}}$across a greased (frictionless) floor, either with$\stackrel{\mathbf{\rightharpoonup }}{\mathbf{F}}$horizontal (case 1) or with tilted upward at an angle$\mathit{\theta }$(case 2). (a)What is the ratio of the refrigerator’s speed in case 2 to its speed in case 1 if you pull for a certain time t? (b) What is this ratio if you pull for a certain distance d?

Force is applied in the horizontal direction in the first case and applied at an angle in the second case.

Newton’s second law states that the net force acting on the object is a product of the mass and acceleration of the object.

Use kinematic equations to find the speed of the refrigerator for a certain time and a certain distance separately.

Also, use angle in the direction of motion, and then we can take the ratio of the speed of the refrigerator in case 2 to the speed in case 1.

Formulae:

$$\begin{gathered} \mathbf{1}\mathbf{.}\mathbf{}{\mathbf{V}}_{\mathbf{f}}\mathbf{=}{\mathbf{V}}_{\mathbf{i}}\mathbf{+}\mathbf{at} \\ \mathbf{2}\mathbf{.}\mathbf{}{\mathbf{V}}_{\mathbf{f}}^{\mathbf{2}}\mathbf{=}{\mathbf{V}}_{\mathbf{i}}^{\mathbf{2}}\mathbf{+}\mathbf{2}\mathbf{a}\mathbf{∆}\mathbf{d} \\ \mathbf{3}\mathbf{.}\mathbf{}\mathbf{}\mathbf{F}\mathbf{=}\mathbf{ma} \end{gathered}$$

When refrigerator starts from rest $V_i=0$.

$V_{f}0+at$

Rearranging for a,

$a=\frac{V_f}{t}$

Using this acceleration in equation of force we get,

$$\begin{gathered} F_x=ma=m\left(\frac{V_f}{t}\right) \\ F\cos \alpha =m\left(\frac{V_f}{t}\right) \\ {V}_f=\frac{F\cos \alpha \left(t\right)}{m} \\ \left(\mathrm{i}\right) \\ \mathrm{Similarly}, \\ {\mathrm{V}}_{\mathrm{f}}^2=0+2\mathrm{a}\left(∆\mathrm{x}\right) \\ \mathrm{a}=\frac{{\mathrm{V}}_{\mathrm{f}}^2}{2\left(∆\mathrm{x}\right)} \end{gathered}$$

(i)

Using a in force equation, we get

$F_x=m\left(\frac{{\mathrm{V}}_{\mathrm{f}}^2}{2\left(∆x\right)}\right)$

Rearranging it for $V_f$,

$$\begin{gathered} {\mathrm{V}}_{\mathrm{f}}^2=Fcos\alpha \frac{2\left(∆x\right)}{m} \\ {\mathrm{V}}_{\mathrm{f}}=\sqrt{\mathrm{Fcos\alpha }\frac{2\left(∆\mathrm{x}\right)}{\mathrm{m}}} \\ \left(\mathrm{ii}\right) \end{gathered}$$

From equation (i) and (ii) we can write ratio,

$\frac{{\mathrm{V}}_{\mathrm{fn}}}{{\mathrm{V}}_{\mathrm{fn}}}=\frac{{\displaystyle \frac{F\cos {\alpha }_2\left(t\right)}{m}}}{\frac{F\cos {\alpha }_1\left(t\right)}{m}}$

For first case,${\alpha }_1=0,\mathrm{and}{\mathrm{\alpha }}_2=\theta$

$$\begin{gathered} \frac{{\mathrm{V}}_{\mathrm{fn}}}{{\mathrm{V}}_{\mathrm{fn}}}=\frac{{\displaystyle \frac{F\cos \theta \left(t\right)}{m}}}{\frac{F\cos \alpha \left(0\right)\left(t\right)}{m}}=\cos \theta \\ \frac{{\mathrm{V}}_{\mathrm{fn}}}{{\mathrm{V}}_{\mathrm{fn}}}=\cos \theta \end{gathered}$$

Therefore, the ratio is equal to $\cos \theta$.

$$\begin{gathered} \frac{{\mathrm{V}}_{\mathrm{fn}}}{{\mathrm{V}}_{\mathrm{fn}}}=\frac{\sqrt{F\cos \theta {\displaystyle \frac{2\left(∆x\right)}{m}}}}{\sqrt{F\cos \left(0\right)\frac{2\left(∆x\right)}{m}}} \\ \frac{{\mathrm{V}}_{\mathrm{fn}}}{{\mathrm{V}}_{\mathrm{fn}}}=\frac{\sqrt{F\cos \theta {\displaystyle \frac{2\left(∆x\right)}{m}}}}{\sqrt{F\cos \left(0\right)\frac{2\left(∆x\right)}{m}}} \\ =\sqrt{\cos \theta } \\ \mathrm{Therefore},\mathrm{the}\mathrm{ratio}\mathrm{is}\mathrm{equal}\mathrm{to}\sqrt{\mathrm{cos\theta }}. \end{gathered}$$