Net Force Physics Problems With Frictional Force and Acceleration

Summary

This educational video script explores physics problems focusing on calculating net force and acceleration. It covers scenarios with varying forces and angles, explaining how to find the net horizontal force and resulting acceleration. The script guides through calculating distance traveled, final speed, and the force exerted by engines, using Newton's laws and kinematic equations. It's an informative resource for understanding force and motion in physics.

Takeaways

• 🔍 The video script focuses on solving physics problems related to calculating net force and acceleration in horizontal directions.
• 📐 The first problem involves a 5 kg box experiencing a 200 N pull and a 60 N frictional force, resulting in a net horizontal force of 140 N.
• 📚 Newton's second law is applied to calculate the acceleration of the box, which is 28 meters per second squared (m/s²).
• 🚀 Using kinematic equations, the script calculates the box's travel distance after 15 seconds, finding it to be 3.15 kilometers.
• 📉 For the second problem, a 12 kg box is pulled by a 350 N force at a 30-degree angle and opposed by a 120 N frictional force.
• 📌 The net horizontal force for the second problem is determined to be 183.1 N by considering the x-component of the applied force.
• ⏱ The acceleration of the second box is calculated to be 15.26 m/s² using the net force and the box's mass.
• 🏎️ The final problem involves a 1200 kg car accelerating from 25 m/s to 60 m/s in 5 seconds, resulting in an acceleration of 7 m/s².
• 🔧 The net force acting on the car is calculated by multiplying its mass by its acceleration, yielding 8400 N.
• 🏗️ The average force exerted by the car's engine is found by adding the net force and the constant frictional force, totaling 11900 N.

Q & A

What is the net horizontal force acting on a 5 kg box when a 200 N force pulls it to the right and a 60 N frictional force opposes it?

The net horizontal force is calculated by subtracting the frictional force from the pulling force. So, it is 200 N - 60 N = 140 N.

How do you calculate the acceleration of the 5 kg box when the net force is 140 N?

According to Newton's second law, acceleration (a) is the net force (F) divided by the mass (m). So, the acceleration is 140 N / 5 kg = 28 m/s².

If the 5 kg box accelerates from rest for 15 seconds with an acceleration of 28 m/s², how far will it travel?

Using the kinematic equation for displacement with initial velocity (v₀) as 0, acceleration (a) as 28 m/s², and time (t) as 15 s, the displacement (d) is calculated as (1/2) a t^2 = (1/2) × 28 × 15^2 = 3150 meters, which is 3.15 kilometers.

What is the net horizontal force on a 12 kg box when a 350 N force at 30 degrees above the horizontal pulls it and a 120 N frictional force opposes it?

The horizontal component of the pulling force is the force multiplied by the cosine of the angle, which is 350 × cos(30°). Subtracting the frictional force gives a net horizontal force of 350 × cos(30°) - 120 N, which is approximately 183.1 N.

What is the acceleration of the 12 kg box when the net horizontal force is 183.1 N?

Using Newton's second law, the acceleration is the net force divided by the mass, which is 183.1 N / 12 kg = 15.26 m/s².

If a 12 kg box is accelerated over a distance of 200 meters, what is its final speed given the acceleration is 15.26 m/s² and the initial speed is zero?

Using the kinematic equation v_final^2 = v_initial^2 + 2ad, and knowing v_initial = 0, a = 15.26 m/s², and d = 200 m, the final speed is √(2 × 15.26 × 200) = 78.1 m/s.

What is the acceleration of a 1200 kg car that speeds up from 25 m/s to 60 m/s in 5 seconds?

Using the equation v_final = v_initial + at, and knowing v_final = 60 m/s, v_initial = 25 m/s, and t = 5 s, the acceleration is (60 - 25) / 5 = 7 m/s².

What is the net force acting on the 1200 kg car that accelerates from 25 m/s to 60 m/s in 5 seconds?

The net force is the mass times the acceleration, which is 1200 kg × 7 m/s² = 8400 N.

If a 1200 kg car experiences a constant frictional force of 3500 N, what is the average force exerted by the engines on the car to achieve the acceleration calculated in the previous question?

The force exerted by the engines is the net force plus the frictional force, which is 8400 N + 3500 N = 11900 N.

How does the direction of frictional force compare to the direction of motion for an object moving to the right?

Friction always opposes motion, so for an object moving to the right, the frictional force will be directed to the left.