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Chapter 11: Problem 1542

An inclined plane making an angle of \(30^{\circ}\) with the horizontal is placed in an uniform electric field \(E=100 \mathrm{Vm}^{-1}\). A particle of mass \(1 \mathrm{~kg}\) and charge \(0.01 \mathrm{c}\) is allowed to slide down from rest from a height of \(1 \mathrm{~m} .\) If the coefficient of friction is \(0.2\) the time taken by the particle to reach the bottom is $\ldots \ldots . .$ sec (A) \(2.337\) (B) \(4.337\) (C) 5 (D) \(1.337\)

Short Answer

Expert verified
The time taken by the particle to reach the bottom of the inclined plane is approximately \(t \approx 1.337 \mathrm{s}\), which corresponds to option (D).

Step by step solution

01

Calculate the Components of Gravitational Force

First, let's identify the gravitational force acting on the particle and separate it into components perpendicular and parallel to the inclined plane. The gravitational force is given by \(F_g = mg\), where \(m = 1 \mathrm{kg}\) and \(g = 9.81 \mathrm{m/s^2}\). We have: - \(F_{g \perp} = mg \cos{30^{\circ}}\) (perpendicular component) - \(F_{g \|} = mg \sin{30^{\circ}}\) (parallel component)
02

Calculate the Electric Force

Now, we need to calculate the electric force acting on the particle due to the electric field \(E\). The electric force is given by \(F_e = qE\), where \(q = 0.01 \mathrm{C}\), and \(E = 100 \mathrm{Vm^{-1}}\). Therefore, the electric force is: \[F_e = (0.01 \mathrm{C})(100 \mathrm{Vm^{-1}}) = 1 \mathrm{N}\] Since the electric field is horizontal and uniform, its component parallel to the inclined plane is: \[F_{e \|} = F_e \sin{30^{\circ}}\]
03

Calculate the Friction Force

The next step is determining the friction force acting on the particle as it slides down the inclined plane. The friction force is given by \(F_f = \mu F_{g\perp}\), where \(\mu\) is the coefficient of friction (\(0.2\) in this case). Thus, \(F_f = \mu (mg \cos{30^{\circ}})\)
04

Calculate the Net Force and Acceleration

Now, we should calculate the net force acting on the particle along the inclined plane and find the acceleration of the particle. The net force is the difference between the parallel components of the electric and gravitational forces (sum of these two forces) and the friction force: \[F_{net} = (F_{g \|} + F_{e \|}) - F_f\] The acceleration of the particle can be found as: \[a = \frac{F_{net}}{m}\]
05

Calculate the Displacement and Time

To calculate the time it takes for the particle to reach the bottom of the inclined plane, we need to determine the displacement along the inclined plane: \[s = \frac{height}{\sin{30^{\circ}}} = \frac{1 \mathrm{m}}{\sin{30^{\circ}}}\] Using one of the equations of motion (\(s = ut + \frac{1}{2} at^2\)), where \(u\) is the initial velocity (zero in this case), and solving for time, we find: \[t = \sqrt{\frac{2s}{a}}\]
06

Plug in Values and Calculate the Answer

Lastly, we plug the values obtained in steps 1 to 5 into the expression for time and calculate the answer: - \(F_{g \perp} = (1 \mathrm{kg})(9.81 \mathrm{m/s^2})\cos{30^{\circ}}\) - \(F_{g \|} = (1 \mathrm{kg})(9.81 \mathrm{m/s^2})\sin{30^{\circ}}\) - \(F_{e \|} = (1 \mathrm{N})\sin{30^{\circ}}\) - \(F_f = (0.2)(1 \mathrm{kg})(9.81 \mathrm{m/s^2})\cos{30^{\circ}}\) - \(F_{net} = (F_{g \|} + F_{e\|}) - F_f\) - \(a = \frac{F_{net}}{1 \mathrm{kg}}\) - \(s = \frac{1 \mathrm{m}}{\sin{30^{\circ}}}\) - \(t = \sqrt{\frac{2s}{a}}\) Calculating these values, we find that the time taken by the particle to reach the bottom is approximately \(t \approx 1.337 \mathrm{s}\), which corresponds to option (D).

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Most popular questions from this chapter

Point charges \(q_{1}=2 \mu c\) and \(q_{2}=-1 \mu c\) care kept at points \(\mathrm{x}=0\) and \(\mathrm{x}=6\) respectively. Electrical potential will be zero at points ..... (A) \(\mathrm{x}=-2, \mathrm{x}=2\) (B) \(\mathrm{x}=1, \mathrm{x}=5\) (C) \(\mathrm{x}=4, \mathrm{x}=12\) (D) \(\mathrm{x}=2, \mathrm{x}=9\)

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The electric Potential \(\mathrm{V}\) at any Point $0(\mathrm{x}, \mathrm{y}, \mathrm{z}\( all in meters \))\( in space is given by \)\mathrm{V}=4 \mathrm{x}^{2}\( volt. The electric field at the point \)(1 \mathrm{~m}, 0.2 \mathrm{~m})\( in volt meter is \)\ldots \ldots .$ (A) 8 , along negative \(\mathrm{x}\) - axis (B) 8 , along positives \(\mathrm{x}\) - axis (C) 16 , along negative \(\mathrm{x}\) -axis (D) 16 , along positives \(\mathrm{x}\) -axis

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Two identical capacitors have the same capacitance \(\mathrm{C}\). one of them is charged to a potential \(\mathrm{V}_{1}\) and the other to \(\mathrm{V}_{2}\). The negative ends of the capacitors are connected together. When the positive ends are also connected, the decrease in energy of the combined system is (A) \((1 / 4) \mathrm{C}\left(\mathrm{V}_{1}^{2}-\mathrm{V}_{2}^{2}\right)\) (B) \((1 / 4) \mathrm{C}\left(\mathrm{V}_{1}^{2}+\mathrm{V}_{2}^{2}\right)\) (C) \((1 / 4) \mathrm{C}\left(\mathrm{V}_{1}-\mathrm{V}_{2}\right)^{2}\) (D) \((1 / 4) \mathrm{C}\left(\mathrm{V}_{1}+\mathrm{V}_{2}\right)^{2}\)
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