Laws of Motion // May 2026

Mastering Work, Energy, and Newton's Second Law

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Understanding the Problem

In this scenario, we are analyzing a $7\text{ kg}$ wagon moving across a frictionless horizontal surface. We are given the following information:

  • Mass ($m$): $7\text{ kg}$
  • Initial velocity ($u$): $4\text{ m/s}$
  • Applied Force ($F$): $10\text{ N}$
  • Distance covered ($d$): $3\text{ m}$

We need to find:

  1. The acceleration ($a$) produced by the force.
  2. The final speed ($v$) of the wagon.

Step 1: Calculating Acceleration

Using Newton's Second Law, which states that the net force applied to an object is equal to the product of its mass and its acceleration ($F = ma$):

$$a = \frac{F}{m}$$ $$a = \frac{10\text{ N}}{7\text{ kg}} \approx 1.43\text{ m/s}^2$$

Step 2: Calculating Final Speed

We can find the final velocity using the kinematic equation of motion, specifically $v^2 = u^2 + 2ad$:

$$v^2 = (4\text{ m/s})^2 + 2 \times (1.43\text{ m/s}^2) \times (3\text{ m})$$ $$v^2 = 16 + 8.58$$ $$v^2 = 24.58$$ $$v = \sqrt{24.58} \approx 4.96\text{ m/s}$$

Alternative Approach: The Work-Energy Theorem

The Work-Energy Theorem states that the work done by a net force equals the change in kinetic energy:

$$W = \Delta KE = \frac{1}{2}mv^2 - \frac{1}{2}mu^2$$

Given $W = F \times d = 10\text{ N} \times 3\text{ m} = 30\text{ J}$:

$$30 = \frac{1}{2} \times 7 \times (v^2 - 16)$$ $$60 = 7(v^2 - 16)$$ $$8.57 = v^2 - 16$$ $$v^2 = 24.57$$ $$v \approx 4.96\text{ m/s}$$

Both methods yield the same result, confirming our calculation is correct.

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