Question

An inclined plane making an angle of 30° with the horizontal is placed in a uniform horizontal electric field $200 \frac{N}{C}$ as shown in the figure. A body of mass 1kg and charge 5 mC is allowed to slide down from rest at a height of 1m. If the coefficient of friction is 0.2, find the time taken by the body to reach the bottom. [g = 9.8 m/s², sin30° = $\frac{1}{2}$; cos30° = $\frac{\sqrt{3}}{2}$]

• A. 0.92 s
• B. 0.46 s
• C. 2.3 s
• D. 1.3 s

Answer 1

Answer: (D)

1.3 s

Answer 2

The forces acting on the body are:

1. Gravitational force: $W = mg$ vertically downwards.
2. Normal force: $N$ perpendicular to the inclined plane, upwards.
3. Electric force: $F_e = qE$ horizontally in the direction of the electric field.
4. Kinetic frictional force: $f_k = \mu N$ opposite to the direction of motion.

Resolving the forces along the inclined plane (downwards is positive) and perpendicular to the inclined plane (upwards is positive):

• Components of gravitational force:

  • Parallel to the incline: $W_{||} = mg \sin \theta$ (downwards).
  • Perpendicular to the incline: $W_{\perp} = mg \cos \theta$ (downwards).

• Components of electric force:

  • Parallel to the incline: $F_{e,||} = qE \cos \theta$ (upwards).
  • Perpendicular to the incline: $F_{e,\perp} = qE \sin \theta$ (downwards).

Along the direction perpendicular to the incline, the net force is zero:

$N - mg \cos \theta - qE \sin \theta = 0$

$N = mg \cos \theta + qE \sin \theta$

Along the direction parallel to the incline (downwards), the net force is $F_{net, ||}$:

$F_{net, ||} = mg \sin \theta - qE \cos \theta - \mu N$

$F_{net, ||} = mg \sin \theta - qE \cos \theta - \mu (mg \cos \theta + qE \sin \theta)$

The acceleration down the incline is $a = \frac{F_{net, ||}}{m}$:

$a = g \sin \theta - \frac{qE}{m} \cos \theta - \mu g \cos \theta - \mu \frac{qE}{m} \sin \theta$

Substitute the given values:

$a = 9.8 \times 0.5 - 1 \times \frac{\sqrt{3}}{2} - 0.2 \times 9.8 \times \frac{\sqrt{3}}{2} - 0.2 \times 1 \times 0.5$

$a = 4.9 - 0.866 - 1.697 - 0.1 = 2.237 \, m/s^2$

The body starts from rest ($u=0$) at a height of $h=1 \, m$. The distance along the inclined plane to reach the bottom is $s = \frac{h}{\sin \theta} = \frac{1}{0.5} = 2 \, m$.

Using the kinematic equation $s = ut + \frac{1}{2}at^2$:

$2 = 0 \times t + \frac{1}{2} \times 2.237 \times t^2$

$t^2 = \frac{2}{1.1185} \approx 1.7881$

$t = \sqrt{1.7881} \approx 1.337 \, s \approx 1.3 \, s$