Answer
The ratio of the refrigerator’s speed in case 2 to its speed in case 1 is $~~\sqrt{cos~\theta}$
Work Step by Step
Case 1: We can find the acceleration:
$F = ma_1$
$a_1 = \frac{F}{m}$
We can find the speed after moving a distance $d$:
$v_1^2 = v_0^2+2a_1d$
$v_1^2 = 2a_1d$
$v_1^2 = \frac{2~F~d}{m}$
$v_1 = \sqrt{\frac{2~F~d}{m}}$
Case 2: We can find the acceleration:
$F~cos~\theta = ma_2$
$a_2 = \frac{F~cos~\theta}{m}$
We can find the speed after moving a distance $d$:
$v_2^2 = v_0^2+2a_2d$
$v_2^2 = 2a_2d$
$v_2^2 = \frac{2~F~d~cos~\theta}{m}$
$v_2 = \sqrt{\frac{2~F~d~cos~\theta}{m}}$
We can find the ratio of $\frac{v_2}{v_1}$:
$\frac{v_2}{v_1} = \frac{\sqrt{\frac{2~F~d~cos~\theta}{m}}}{\sqrt{\frac{2~F~d}{m}}}$
$\frac{v_2}{v_1} = \sqrt{cos~\theta}$
The ratio of the refrigerator’s speed in case 2 to its speed in case 1 is $~~\sqrt{cos~\theta}$
