Force on Particle with Velocity v Equal to 4x

❓ Question

An object with mass 500 g moves along the x-axis with speed

$$v=4\sqrt{x}\text{ m/s}\mathrm{.}$$

The force acting on the object is: ?

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🖼️ Question Image

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✍️ Short Explanation

This problem is based on:

👉 Newton’s Second Law
👉 Variable velocity
👉 Relation between velocity and acceleration.

Main idea:

When velocity depends on position:

$$\boxed{ a=v\frac{dv}{dx} }$$

Then use:

$$\boxed{ F=ma }$$

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🔷 Step 1 — Given Data 💯

Mass:

$$m=500\text{ g}=0.5\text{ kg}$$

Velocity:

$$v=4\sqrt{x}$$

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🔷 Step 2 — Differentiate Velocity

Use:

$$a=v\frac{dv}{dx}$$

First find:

$$\frac{dv}{dx}$$

Given:

$$v=4x^{1/2}$$

Differentiate:

$$\frac{dv}{dx} = 4\cdot\frac12 x^{-1/2}$$
$$= \frac{2}{\sqrt{x}}$$

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🔷 Step 3 — Find Acceleration

Now:

$$a=v\frac{dv}{dx}$$

Substitute:

$$a= (4\sqrt{x}) \left( \frac{2}{\sqrt{x}} \right)$$
$$=8\text{ m/s}^2$$

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🔷 Step 4 — Find Force

Using:

$$F=ma$$$$F=0.5\times8$$
$$=4\text{ N}$$

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🔷 Step 5 — JEE Trap Alert 🚨

❌ Directly $a=\frac{dv}{dt}$ use karna

❌ Mass conversion $500g=0.5kg$ bhool jaana

❌ Chain rule miss kar dena

Remember:

$$\boxed{ a=v\frac{dv}{dx} }$$

when velocity is given as function of position.

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✅ Final Answer

$$\boxed{ 4\text{ N} }$$

(Option 4)

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