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Chapter 5: Problem 602

A meter stick of mass \(400 \mathrm{gm}\) is pivoted at one end and displaced through an angle 600 the increase in its P.E. is \(\overline{\\{\mathrm{A}\\} 2}\) \(\\{B\\} 3\) \(\\{\) C \(\\}\) Zero \(\\{\mathrm{D}\\} 1\)

Short Answer

Expert verified
The increase in potential energy of the meter stick is approximately 1.703 J, which is closest to \(\overline{\mathrm{A}} 2\).

Step by step solution

01

Convert mass to kg and angle to radians

Firstly, convert the mass of the meter stick from grams to kilograms: \( mass = \dfrac{400~\mathrm{grams}}{1000} = 0.4 ~\mathrm{kg} \) Now convert the angle of displacement from 60 degrees to radians: \( \theta = 60^\circ * \frac{\pi}{180^\circ} = \frac{\pi}{3} ~\mathrm{radians} \)
02

Calculate the change in height

We need to find out the change in height of the meter stick's center of mass due to pivoting. When it is horizontal, the center of mass is right at the middle of the stick (0.5 meters). When it pivots, it moves upwards to form a 30-60-90 degree triangle. The height of the triangle is given by: \( h = 0.5 \times \sin{(\pi/3)} \)
03

Calculate the increase in potential energy

To find the increase in potential energy (ΔP.E.), we need to multiply the mass, gravitational acceleration (g = 9.81 m/s^2), and the change in height: ΔP.E. = mass × g × h ΔP.E. = 0.4 × 9.81 × (0.5 × \(\sin{(π/3)}\)) ΔP.E. ≈ 1.703 J (approx)
04

Choose the correct option

Now that we have calculated the increase in potential energy (ΔP.E.), we can choose the correct option among the given choices. The answer is closest to 2, so the correct option is: \(\overline{\mathrm{A}} 2\)

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