Calculating Frictional Force and Acceleration for a Box Being Pulled Along a Horizontal Plane

Objective: This article explains the process of calculating the frictional force and acceleration of a 10 kg box when a force of 40.0 N is applied at an angle of 30° on a horizontal surface. It covers step-by-step calculations and provides a comprehensive solution to the problem, suitable for students, educators, and those interested in physics and mechanics.

Step-by-Step Guide to Calculating Frictional Force and Acceleration

Problem Context: A 10 kg box is being pulled along a horizontal plane by a force of 40.0 N applied at an angle of 30°. Assuming a coefficient of kinetic friction of 0.3, the goal is to determine the frictional force and the acceleration of the box.

Step 1: Calculate the Normal Force

The first step is to understand the forces at play. The weight of the box (W) is given by:

W mg 10 kg times; 9.81 m/s^2 98.1 N

The applied force (F) has both vertical and horizontal components:

F_vertical F times; sin(30°) 40.0 N times; 0.5 20.0 N

The normal force (N) is the weight of the box minus the vertical component of the applied force:

N W - F_vertical 98.1 N - 20.0 N 78.1 N

Step 2: Calculate the Frictional Force

The frictional force (F_friction) can be calculated using the normal force (N) and the coefficient of kinetic friction (μ):

F_friction μN 0.3 times; 78.1 N 23.43 N

Step 3: Calculate the Horizontal Component of the Applied Force

The horizontal component of the applied force (F_horizontal) is:

F_horizontal F times; cos(30°) 40.0 N times; frac{sqrt{3}}{2} ≈ 34.64 N

Step 4: Calculate the Net Force

The net force (F_net) acting on the box is the horizontal component of the applied force minus the frictional force:

F_net F_horizontal - F_friction 34.64 N - 23.43 N 11.21 N

Step 5: Calculate the Acceleration

Using Newton's second law, we find the acceleration (a) of the box:

F_net ma implies a frac{F_{net}}{m} frac{11.21 N}{10 kg} ≈ 1.12 m/s^2

Summary of Results

Frictional Force: ≈ 23.43 N

Acceleration of the Box: ≈ 1.12 m/s^2

Revised Calculation

Alternatively, another approach is to consider the force of kinetic friction as:

The force of kinetic friction is opposing the x-component of the 40.0 N force, which is equal to -40.0 N times; 0.3 times; cos(30°) -10.4 N.

The acceleration of the box is equal to the net force divided by the mass of the box. The net force is equal to the x-component of the force minus the force of friction:

F_net 40.0 N times; cos(30°) - 10.4 N 34.64 N - 10.4 N 24.2 N

Solving for the acceleration of the box:

acceleration F_{net} / m frac{24.2 N}{40 kg} ≈ 0.61 m/s^2

As a Summary: The frictional force is 10.4 N and the acceleration of the box is equal to 0.61 m/s^2. The discrepancy between the two calculations is due to differences in the mass and the given conditions.