An inclined plane making an angle of 30 degrees with the horizontal is placed in a uniform horizonta

Q: Why does the electric field have a component perpendicular to the incline?

Because the electric field is horizontal, and the incline is at an angle of $30^\circ$ to the horizontal. The electric force vector is not parallel to the incline, so it must be resolved into a component pushing *up* the incline and a component pushing *into* the incline.

Q: Why do we add the electric force component to the normal force?

The normal force must balance all forces pressing the block into the surface. Since gravity pulls it down and the horizontal electric field pushes it into the incline, both contribute to the total normal force.

Q: How do we know the direction of the friction force?

Kinetic friction always opposes the direction of relative motion. Since the block is sliding *down* the incline, the friction force must point *up* the incline.

Q: Why is the distance traveled $2\text{ m}$ and not $1\text{ m}$?

The block slides along the hypotenuse of the inclined plane, not straight down. Using trigonometry, the length of the incline $s$ is related to the height $h$ by $\sin 30^\circ = h/s$. Since $h = 1\text{ m}$, $s = 1 / 0.5 = 2\text{ m}$.

Q: What if the electric field was pointing to the left instead of the right?

If the electric field pointed left, it would pull the block *away* from the incline (reducing the normal force and friction) and pull it *down* the incline (increasing the acceleration).

Question

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

Options:

(A) $0.92	ext{ s}$

(B) $0.46	ext{ s}$

(C) $2.3	ext{ s}$

(D) $1.3	ext{ s}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing next to an inclined plane, watching a charged block slide down. This isn't just a simple gravity problem; there is a horizontal electric field pushing against the block! To solve this, we must break down every force acting on the block into components that are parallel and perpendicular to the incline.

First, we have gravity ($mg$) pulling straight down. When resolved, it gives us a component pulling the block down the incline ($mg \sin 30^\circ$) and a component pressing it into the surface ($mg \cos 30^\circ$).

Next, we have the horizontal electric field ($E$). Because the field is horizontal and the incline is angled, the electric force ($qE$) also has two components. It pushes the block *up* the incline ($qE \cos 30^\circ$) and presses it *into* the incline ($qE \sin 30^\circ$).

### The Master Equation

Before we can find the acceleration, we need to know the kinetic friction. Friction depends on the normal force ($N$), which must balance all the forces pressing into the surface.

$$N = mg \cos 30^\circ + qE \sin 30^\circ$$

Plugging in our values, gravity contributes about $8.49	ext{ N}$, and the electric field adds exactly $0.5	ext{ N}$. This gives us a total normal force of approximately $9	ext{ N}$. With a coefficient of friction $\mu = 0.2$, the kinetic friction opposing the motion is:

$$f = \mu N = 0.2 	imes 9 = 1.8	ext{ N}$$

Now, we apply Newton's Second Law along the incline. The net force driving the block down is the gravitational component minus the friction and the electric field component:

$$F_{	ext{net}} = mg \sin 30^\circ - f - qE \cos 30^\circ$$

Substituting the numbers, we get $4.9	ext{ N} - 1.8	ext{ N} - 0.866	ext{ N}$, which results in a net force (and thus an acceleration, since $m = 1	ext{ kg}$) of $a = 2.234	ext{ m/s}^2$.

### Final Calculation

To find the time it takes to reach the bottom, we need the total distance $s$ the block travels along the incline. Using basic trigonometry with the given height $h = 1	ext{ m}$:

$$s = \frac{h}{\sin 30^\circ} = \frac{1}{0.5} = 2	ext{ m}$$

Finally, we use the kinematic equation for an object starting from rest:

$$s = ut + \frac{1}{2}at^2$$

$$2 = 0 + \frac{1}{2}(2.234)t^2$$

Solving for $t^2$ gives us approximately $1.79$, which means the time $t$ is about **$1.33	ext{ s}$**. Looking at our options, $1.3	ext{ s}$ is the perfect match!

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Visualized Solution

$Free Body Diagram$

$Resolving Forces$

$Normal Force Equilibrium$

$Calculating Normal Force$

$Kinetic Friction$

$Equation of Motion$

$Substituting Values$

$Calculating Acceleration$

$Distance Along Incline$

$Time Taken$

The Sigma Insight: Static and Kinetic Friction

Analyzing the Setup

The Master Equation

Final Calculation

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Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Free Body Diagram

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Resolving Forces

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Normal Force Equilibrium

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculating Normal Force

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Kinetic Friction

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Equation of Motion

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Substituting Values

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculating Acceleration

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Distance Along Incline

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Time Taken