Physical Setting / Physics Regents Examinations: Solved Mechanics Problems Using Equations

Using the Equations to Solve Physics Problems
This guide contains simple examples that use each of the equations given in Reference Table. On the Regents examination in physics, you may be required to rearrange these equations or apply more than one equation in order to solve a problem.

MECHANICS 

(Note: Solved examples below are each preceded by a mechanics equation / formula)

ME1.

EXAMPLE: It takes a car 0.20 hour to travel a distance of 12 kilometers.

ME2.

EXAMPLE: An object is uniformly accelerated from rest to a speed of 25 meters per second in 10 seconds.

ME3. vf = vi + at
EXAMPLE: An object moving with an initial velocity of 3.0 meters per second accelerates for 15 seconds at 2.0 meters per second. What is the final velocity of the object?

ME4.  

EXAMPLE: A cart with an initial velocity of 6.00 meters per second rolls down an inclined plane with an acceleration of 4.00 meters per second squared. What distance will it travel in 3.00 seconds?

ME5.

EXAMPLE: A car moving at a constant speed of 4.10 meters per second then accelerates uniformly at 3.20 meters per second squared. What will be its speed when it has traveled a distance of 40.0 meters?

ME6. Ay = A sin θ

ME7. Ax = A cos θ
EXAMPLE: A projectile is fired at an angle of 30° to the horizontal with an initial velocity of 100 meters per second. Calculate the vertical and horizontal components of the projectile’s initial velocity.

ME8.

EXAMPLE: A 3.0-kilogram mass is being moved along a horizontal surface by a force of 6.0 newtons. If the surface is frictionless, what is the acceleration produced by the 6.0-newton force?

ME9.

EXAMPLE: A 10-newton rubber object is dragged across a dry, horizontal concrete floor at constant speed. Calculate the force of friction on the object.
Since the object is on a horizontal surface, the normal force is equal to its weight (10 N). Use Reference Table C to obtain the kinetic coefficient of friction between rubber and dry concrete (0.68).

ME10.

EXAMPLE: Two masses of 10.0 kg and 1.0 kg, respectively, are located 1.0 meter apart. How large a gravitational force does each mass exert on the other?

ME11.

EXAMPLE: The acceleration due to gravity on planet A is 20 meters per second squared. On this planet what is the gravitational force on an object whose mass is 2.0 kilograms?

ME12. p = mv
EXAMPLE: A cart whose mass is 20 kg moves with a velocity of 5.0 meters per second eastward. What is its momentum?

ME13.

EXAMPLE: A 5.0-kilogram object traveling at 5.0 meters per second east collides with a 2.0-kilogram object traveling at 2.0 meters per second east. After the collision, the 2.0-kilogram object is moving at 4.0 meters per second east. What is the velocity of the 5.0-kilogram object? [Neglect friction.]
Designate east by using a positive sign.

ME14.

EXAMPLE: A spring exerts a force of 50.0 newtons on a cart located on a frictionless plane. The cart has a mass of 2.0 kilograms and the force acts for 0.20 second.

The change of momentum of the cart is also 10 N • s, and that is the same thing as 10 kilogram-meters per second.

ME15. Fs = kx
EXAMPLE: A block is suspended from a spring which has a spring constant, k, of 200 newtons per meter. What is the force on the spring if the spring is stretched 0.20 meter?

ME16.

EXAMPLE: A block is suspended from a spring and as a result the spring is stretched 0.20 meter. The spring constant is 200 newtons per meter. How much potential energy is stored in the spring?

ME17.

ME18.

EXAMPLE: An object moves around a circle whose radius is 2.0 meters. The constant speed of the object is 6.0 meters per second and the mass of the object is 0.20 kilogram.

ME19.

EXAMPLE: A 3.0-kilogram mass is raised 4.0 meters from a surface. Calculate its gain in gravitational potential energy.

ME20.

EXAMPLE: A car with a mass of 1000 kilograms travels with a speed of 20 meters per second.

ME21.

EXAMPLE: If a horizontal force of 30 newtons is used to push an object 40 meters along a horizontal surface, the work done on the object

 

Note: 1 newton-meter = 1 joule

ME22.

EXAMPLE: As an object, initially at rest, slides down a plane, it loses 25 joules of potential energy. If the object gains 18 joules of kinetic energy, what is the change in the internal energy of the object? How is the internal energy change used?
At the top of the incline, the total energy of the object is present as potential energy and internal energy. At the bottom of the incline, the lost potential energy is converted to kinetic energy and additional internal energy. The total energy of the object remains constant throughout the slide. Modify equation ME22 to read:

The change in the internal energy is used to do work against the friction between the object and the incline.

ME23.

Power is the rate of doing work.
 

EXAMPLE: An object whose mass is 3.0 kilograms is moved at constant speed by a force of 5.0 newtons. If the object is moved 40 meters in 20 seconds, then

But the work which is done is equal to the product of the applied force and the distance that the object is moved: