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Chapter 4: Problem 93

In the physics laboratory, a glider is released from rest on a frictionless air track inclined at an angle. If the glider has gained a speed of $25.0 \mathrm{cm} / \mathrm{s}\( in traveling \)50.0 \mathrm{cm}$ from the starting point, what was the angle of inclination of the track? Draw a graph of \(v_{x}(t)\) when the positive \(x\) -axis points down the track.

Short Answer

Expert verified
The angle of inclination of the track is approximately \(0.447^{\circ}\). The graph of \(v_{x}(t)\) is a straight line starting at the origin, with a slope of \(0.125\).

Step by step solution

01

Convert cm to m and setup known variables

To make the calculations easier, convert the given values into meters: Initial speed \((u) = 0\) (since the glider is released from rest) Final speed \((v) = 25.0cm/s = 0.25m/s\) Distance \((s) = 50.0cm = 0.5m\)
02

Find the acceleration along the track

We can use the equation of motion (which applies to constant acceleration) to find the acceleration along the track: $$ v^{2} = u^{2} + 2as $$ Plug in the known values and solve for \(a\): $$ (0.25)^{2} = 0^{2} + 2a(0.5) $$ $$ a = 0.125 m/s^{2} $$
03

Determine the gravitational force component along the track

Since the glider experiences gravity and the only force acting on it is the gravitational force component along the track, we have: $$ F = ma $$ $$ mg \sin\theta = ma $$ Where \(g\) is the gravitational acceleration \((9.81 m/s^{2})\) and \(\theta\) is the angle of inclination. We can now solve for \(\theta\): $$ \sin\theta = \frac{a}{g} $$
04

Calculate the angle of inclination

Plug in the values of \(a\) and \(g\) and solve for \(\theta\): $$ \sin\theta = \frac{0.125}{9.81} $$ $$ \theta = \arcsin(\frac{0.125}{9.81}) $$ $$ \theta \approx 0.447^{\circ} $$
05

Draw a graph of \(v_{x}(t)\) along the track

Use the equation of motion along the x-axis to express the final speed as a function of time for a positive x-axis pointing down the track: $$ v_{x}(t) = u + at $$ As the glider starts from rest, we have: $$ v_{x}(t) = at $$ To plot \(v_{x}(t)\), just remember that it's a straight line with the slope equal to the acceleration component along the x-axis \((a)\). The line starts at the origin (\((0,0)\)) and has a slope of \(0.125\). In summary, the angle of inclination of the track is approximately \(0.447^{\circ}\), and the graph of \(v_{x}(t)\) is a straight line starting at the origin and with a slope of \(0.125\).

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