A constant horizontal force acts on an object of mass 8 kg for 3 seconds, increasing its velocity uniformly from 4 m/s to 6 m/s. What is the magnitude of this applied force?

Introduction / Context:
Newton second law of motion provides a relation between force, mass, and acceleration. Whenever a constant resultant force acts on a mass, it produces uniform acceleration and changes the velocity of the object. This question asks you to determine the magnitude of such a force when the mass, initial velocity, final velocity, and time interval are known. This is a standard type of problem in mechanics and is directly based on the formula F = m * a, along with the definition of acceleration as change in velocity per unit time.

Given Data / Assumptions:
• Mass of the object m = 8 kg. • Initial velocity u = 4 m/s. • Final velocity v = 6 m/s. • Time interval t = 3 s. • The applied force is constant, so acceleration is uniform. • Motion is taken along a straight line and only one significant horizontal force is considered.

Concept / Approach:
First, we calculate the acceleration using the kinematic relation a = (v - u) / t. Once we know the acceleration, we apply Newton second law in the form F = m * a to find the magnitude of the net force acting on the object. The problem does not ask for direction or sign; it simply needs the numerical magnitude, so we will work with positive values. Accurate substitution of data and careful handling of fractions are important to obtain the correct decimal value.

Step-by-Step Solution:
Step 1: Use the definition of acceleration a = (v - u) / t. Step 2: Substitute v = 6 m/s, u = 4 m/s, and t = 3 s. Step 3: Compute a = (6 - 4) / 3 = 2 / 3 m/s^2. Step 4: Apply Newton second law, F = m * a. Step 5: Substitute m = 8 kg and a = 2 / 3 m/s^2. Step 6: Calculate F = 8 * (2 / 3) = 16 / 3 N. Step 7: Convert 16 / 3 into decimal form: 16 / 3 ≈ 5.33 N.

Verification / Alternative check:
We can check the order of magnitude. An acceleration of roughly 0.67 m/s^2 acting on 8 kg should give a force slightly greater than 5 N because 8 * 0.5 is 4 and 8 * 1 is 8, so 8 * 0.67 should be between 4 and 8. The calculated value 5.33 N fits well within this range. Also, the small change in velocity of only 2 m/s over 3 seconds suggests a modest acceleration, so a modest force is expected, which agrees with the answer.

Why Other Options Are Wrong:
Option b (4.33 N): This value is smaller than 16 / 3 N and arises if someone uses an incorrect acceleration or miscalculates the product. Option c (6.33 N): This is larger than the correct answer and may result from taking a = 3 / 2 instead of 2 / 3 or from a numerical error. Option d (3.33 N): This corresponds to using a smaller acceleration, such as 1 / 3 m/s^2, which does not match the speed change given. Option e (2.67 N): This is about half of the correct value and may come from incorrectly using half of the actual mass or other algebraic mistakes.

Common Pitfalls:
Learners sometimes mix up the formula for acceleration, writing (u - v) / t instead of (v - u) / t, which changes the sign. Others may forget that the time is 3 seconds and mistakenly divide by a different number. Rounding errors can also lead to answers like 5.3 N or 5.4 N, which may not exactly match the given options. To avoid mistakes, always compute the difference in velocities carefully, divide correctly by time, and then multiply by mass with attention to fraction handling.

Final Answer:
The magnitude of the applied force is 5.33 N.