Question

An object starts from rest and slides with negligible friction down an air track tipped at an angle theta from the horizontal. A student records values of the object's position along the track at various times. the value of theta can best be determined from which of the following? the y-intercept of a graph of position as a function of time the y-intercept of a graph of position as a function of the square of time the slope of a graph of position as a function of time the slope of a graph of position as a function of the square of time

Answer 1

Answer:

The value of θ can be determined by slope of graph as function of square of time.

Explanation:

Lets take

Mass = m

Angle of inclination =θ

From diagram

mg sinθ = m a

So

a=g sinθ

We know that

$S=ut+\dfrac{1}{2}at^2$

Given that block is in rest position initially.

u= 0

$S=\dfrac{1}{2}(gsin\theta ) t^2$

We can say that

S∝ t²      If we assume that t and x and S as y y= m x

$S=\dfrac{1}{2}(gsin\theta ) t^2$

Here slope ,m

$m=\dfrac{1}{2}(gsin\theta )$

So the value of θ can be determined by slope of graph as function of square of time.

Answer 2

The value of theta can best be determined from the slope of a graph of position as a function of the square of time.

Let the angle of inclination of the track, = θ

Let the acceleration of the object = a

Generally, the slope of a graph is given as;

$tan(\theta) = \frac{\Delta y}{\Delta x}$

The vertical component of the force on the object along the track is calculated as;

$Fcos(\theta) = mg$

The horizontal component of the force on the object along the track is calculated as;

$Fsin(\theta) = ma$

The coefficient of friction on the object is calculated as;

$\mu_k = \frac{Fsin(\theta)}{Fcos(\theta)} = \frac{ma}{mg} \\\\ tan(\theta)= \frac{a}{g} \\\\ tan(\theta)= \frac{dv/dt}{g} \\\\v = \frac{dx}{dt} \\\\tan(\theta) = \frac{dx^2/d^2t}{g} \\\\tan(\theta) = \frac{dx^2}{d^2t\ \times \ g }$

Thus, the value of theta can best be determined from the slope of a graph of position as a function of the square of time.

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