Practice Applying Force & Acceleration Formulas

Practice Applying Force & Acceleration Formulas
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  • 0:04 Newton's Force Laws
  • 0:52 Sample Probles
  • 6:38 Lesson Summary
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Lesson Transcript
Instructor: Michael Blosser

Michael has a Masters in Physics and a Masters in International Development. He has over 5 years of teaching experience, teaching Physics, Math, and English classes.

In this lesson, we will introduce force and acceleration formulas using Newton's Force Laws and demonstrate kinematic equations, including examples, to solve real-world problems.

Newton's Force Laws

Isaac Newton created his revolutionary three laws of motion in the 17th century. One of the most influential of Newton's laws of motion was Newton's Second Law, which states that the net force on an object is equal to the mass of the object times the acceleration of the object.

This law can be expressed as an equation:

Force = Mass x Acceleration

Or, more simply:

F = ma

Using Newton's three laws of motion and the kinematic equations, or equations that can be used to describe an object's motion, we can solve for numerous unknown forces and other variables of an object in real-world situations. The kinematic equation used in this lesson is:

Kinematic Equation

Sample Problems

  • A 250 kg car is driving with a constant acceleration of 50 m/s2. Calculate the force required to produce this acceleration. Also, find the new acceleration of the car if the force accelerating the car dropped to 5000 Newtons.

In this equation, a car with a given mass is accelerating. Therefore, we know we should use Newton's Second Law to solve this problem.

Newton's Second Law as a formula is:

F = m * a

For the first part of the problem, we need to find the force that produced the acceleration by plugging in the values:

F = m * a = 250 * 50

This yields a force of 12,500 Newtons. Additionally, the question asks us to solve for the new acceleration of the car if the force drops to 5000 N.

Therefore, using the same equation:

F = m * a

a = F/m

a = 5000 / 250

This gives us a new acceleration of 20 m/s2.

Let's try another one.

  • A man is pulling a 30 kg box across a floor with a rope at an angle 30°, with an applied force Fa = 500 N. The force applied by the man causes the box to accelerate across the level surface of floor, which has a coefficient of kinetic friction μ, = 0.5. What's the acceleration of the box?

To solve this problem we'll need to draw a free body diagram and analyze the forces in both the x and y directions. Note that the acceleration of the box will be in the x direction; because the box slides across the level surface of the floor, we know that it has no motion in the y direction, only the x direction. Therefore, acceleration in the y direction is 0, and we know that the net force in the y direction must also be equal to 0.

Let's start our analysis with a free body diagram. From the free body diagram we can use sine and cosine to solve for the x and y components of the applied force vector. The applied force vector, FA, equals the x component, FAX plus the y component, FAY, which equals the magnitude of the applied force times the cosine of theta in the x direction plus the magnitude of the applied force times the sine of theta in the y direction.

Now that we've eliminated a few unknowns, let's analyze the forces in the x and y directions using Newton's Second Law. First let's solve for the only unknown force in the y direction - the normal force.

The net force in the y direction equals the normal force plus the y component of the applied force FAY minus the force of gravity, Fg. Velocity in the y direction equals 0; therefore, we know the net force in the y direction also equals 0. Solving for the normal force, we get:

FN = Fg - FAY

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