Work B1.4 and Energy.

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Presentation transcript:

Work B1.4 and Energy

Force A force can be described as a or a . The unit for force is the newton. symbol: N We will only review forces briefly and qualitatively, but the formula for calculating forces is: force = mass × acceleration (N) (m/s2) (kg) Sir Isaac Newton (1642 – 1727) “One newton equals one kilogram metre per second squared.”

Balanced Forces When the forces acting on an object are balanced, it will remain at a constant velocity (including zero). Force 1 and force 2 are balanced (equal in magnitude and opposite in direction). To lift a ball up at a constant speed, you must apply a force that is equal to the force that gravity pulls on the ball with (i.e. the weight of the object).

Unbalanced Forces If the forces acting on an object are unbalanced, then the object will accelerate. The direction of the acceleration will be in the direction of the strongest force. Forces can be used to transfer energy, or convert energy from one form to another.

In physics, we speak about work being done on an object. Work & Energy Work and energy are two concepts which are very closely related to one another. energy: the ability to do work work: the transfer of energy from one object to another, or from one form of energy to another. In physics, we speak about work being done on an object. W = work (J) = applied force (N) = distance the object moves (m)

The S.I. unit for work is the joule. 1 joule = 1 J The joule, like the Newton, is a derived unit. To see how we break it down from it’s fundamental units, we look at the formula: James Prescott Joule (1818 – 1889) “One joule is equal to one newton-metre.” (J) (m) (N) The joule broken down into its fundamental units. So, the unit that we use for energy is also the joule. Work is a transfer (or conversion) of energy.

These directions are perpendicular! The direction the object moves must be in the same direction as the applied force, otherwise no work is being done. For each of the following examples, do I do any work? 1) I carry a 2.0 kg wheel of cheese 12.0 m across the room. The force that I apply to hold the cheese up is vertical, and the cheese only moves horizontally. Stewie sez: These directions are perpendicular!

2) I lift a 2.0 kg wheel of cheese from the ground to the top of a table. The direction of the displacement is the same as the direction as the applied force. 3) I push against a wall with a force of 130 N for 10.0 s. The wall did not move.

examples: Practice Problems p. 160 18) A tugboat is towing a tanker through a canal using a towrope. Calculate the work done by the tugboat if it applies an average horizontal force of 6.50 x 103 N on the towrope while towing the tanker through a horizontal distance of 150 m. 9.75 x 105 J 19) A large crane did 2.2 x 104 J of work in lifting a demolition ball a vertical distance of 9.5 m. Calculate the average force exerted by the chain on the ball. 2.3 x 103 N 20) A large crane does 2.2 x 104 J of work in lifting an object. How much energy is gained by the object? 2.2 x 104 J

Work Input and Work Output The work (or energy) input can be calculated using the formula W = Fd. When doing work, sometimes some energy is lost as useless heat because of a resistive force, like friction. The work (or energy) output is the amount of energy the object or system gains as a result of the work being done. It is the work input minus any energy lost as the result of friction.

Calculating Work From Force vs Displacement Graphs To calculate work from a force-displacement graph, we calculate THE AREA UNDER THE CURVE. By “under the curve,” we mean the area between the line and the x-axis. Force (N) 5.0 Let’s say we wanted to calculate the work done on the object in moving it 10.0 m. The area under the curve forms a rectangle. area = length x width 10.0 = (5.0 N) (10.0 m) Displacement (m) = 50 J So, the work done was 50 J.

Homework: read pages 155 – 160 B1.4 Check and Reflect Prepare for section 1 quiz Section Review page 162-163 # 1-24