class 051003

Text Homework:  Read sections 1-6 of Chapter 4.

What is the acceleration of a 5 kg object on which the net force is 10 Newtons?

a = F_net / m = 10 Newtons / (5 kg) = 2 Newtons / kg

= 2 kg m/s^2 / kg = 2 m/s^2.

How much net force is required to acceleration a 2500 kg vehicle at 2 m/s^2?

F_net = m a = 2500 kg * 2 m/s^2 = 5000 kg m/s^2 = 5000 N.

What is the kinetic energy of a 110 kg running back moving at 7 meters / second?

KE = .5 m v^2 = .5 * 110 kg * (7 m/s)^2 = 2725 kg m^2 / s^2 = 2725 Joules.

Note that kg m^2 / s^2 = kg m/s^2 * m = N * m = Joules.

The Joule is the unit of work/energy.  Basic definition of work is F * `ds, which clearly has units of Newtons * meters.

If friction exerts a force of 30 N in the direction opposite that of motion on a mass of 80 kg, while I pull on the object with a force of 100 N, then if the object is intially at rest:

• What will be the acceleration of the object?
• The net force on the object will be 100 N - 30 N = 70 N.  So its acceleration will by 70 N / (80 kg) = .88 N / kg = .88 kg m/s^2 / kg = .88 m/s^2.

• If I pull on the above object through a displacement of 3 meters, what velocity will it attain?
• Using v0 = 0, a = .88 m/s^2 and `ds = 3 m we use the fourth equation of motion to get vf = +- 2.3 m/s, approx.. 
• Rejecting the -2.3 m/s we find that vf = 2.3 m/s.

• What then will be the kinetic energy of the object at the end of the 3 meters?
• The KE of the object will be .5 m v^2 = .5 * 80 kg * (2.3 m/s)^2 = 210 Joules, approx..

Recall the basic work-energy theorem:

`dW_net = `d(KE).

What is `dW_net in the above problem?

`dW_net = F_net * `ds = 70 N * 3 m = 210 N m = 210 J.

We found above that `dKE = 210 J.

This confirms the work-energy theorem.

The calculation of `dKE was based on F_net = m a and the equations of uniformly accelerated motion.  The result was to be expected, because the work-energy theorem was derived symbolically, based on the equations of uniformly accelerated motion and Newton's Second Law.

What forces act on a ball rolling down an incline?

The gravitational force acts straight down.

A normal force (an elastic reaction of the incline to the compression or bending that results from the weight of the object) acts perpendicular to the incline.

A frictional force acts in the direction opposite motion.

The resultant of the normal and gravitational forces is the diagonal of the parallelogram formed by the two force vectors.  The normal force is such that this resultant acts down the incline.

The net force is therefore in the direction along the incline.