Question

A friend claims that her car can accelerate from a stop to 60 mi/h (26.8 m/s) in 5.1 s , but the speedometer is broken. You decide to test her claim by riding with her, and you bring along a small metal washer, a short length of string, a protractor, and a pen that can write on glass. Sitting in the passenger seat, you tie the washer to one end of the string and then let the string-washer combination hang straight down by rolling up the passenger-door window to pinch the free end of the string. While the car is stationary, you draw a line on the window alongside the vertical string.
How large is the angle the string makes with the line you drew if the car accelerates at the rate your friend claims?

If an identical washer does not slide on the horizontal dashboard while this is happening, what is the ratio of the frictional force to the gravitational force exerted on this washer?

Answer 1

I attached a diagram, according to the description given, to better understand the problem. The car accelerates, at a 'a' speed, which will also experience the hanging object, in two components:

The one in x, with 'a' in the opposite direction to the car's address

The one found in y, the product of gravity. The two components are related through the tangent and the respective angle, as well,

$tan\theta = \frac{ma}{mg}$

$tan\theta = \frac{a}{g}$

To identify, we know that it can be expressed as a function of speed,

$a= \frac{\Delta v}{t} = \frac{v-u_0}{5.1}\frac{26.8}{5.1}$

$a=5.26m/s^2$

Replacing in our angle formula,

$tan\theta = \frac{5.26}{9.8}$

$tan\theta = 0.54$

$\theta = tan^{-1}(0.54)$

$\theta = 28.4\°$

If there is no friction, we consider the vertical forces of both the Washes and the Normal car. So,

$F=ma$

Therefore the relationship with the vertical component would be given by,

$\Rightarrow \frac{F}{mg} = \frac{ma}{mg} = \frac{a}{g}=\frac{5.26}{9.8}=0.54 =54\%$