7.5 - Work

When the force acting on an object is constant, work can be described without calculus But constant force is very limiting. Take a simple spring. → No force needed → Constant force to hold at 3 → Variable force to keep compressing

Work Done by a Variable Force Where F(x) is a continuously varying force The units of work are: foot-pound, meter-ton, newton-meters, ergs, or Joules Formulas Force on a springGravitational force between 2 objects Attraction force between 2 objects in a vacuum

Review: Hooke’s Law: A spring has a natural length of 1 m. A force of 24 N stretches the spring to 1.8 m. a Find k : b How much work would be needed to stretch the spring 3m beyond its natural length?

Example – Gravity Determine the work in propelling a 4 metric-ton spacecraft 200 miles above earth. (Use 4000 miles for the radius of Earth) Simplify the formula: Use initial conditions: Find work: mile-tons Answer

Liquid Problems Change from a constant force to varying force

5 ft 10 ft 4 ft I want to pump the water out of this tank. How much work is done? Where F is the weight of the water and the distance is from the bottom to top

5 ft 10 ft 4 ft 5 ft 10 ft ft dx I want to pump the water out of this tank. How much work is done? forcedistance

5 ft 10 ft 4 ft 5 ft 10 ft ft dx I want to pump the water out of this tank. How much work is done? A 1 horsepower pump, rated at 550 ft-lb/sec, could empty the tank in about 14 minutes.

10 ft 2 ft 10 ft A conical tank is filled to within 2 ft of the top with fuel weighing 57 lb/ft 3. How much work was required to pump in the fuel from the top? Consider one slice (slab) first:

10 ft 2 ft 10 ft A conical tank if filled to within 2 ft of the top with fuel weighing 57 lb/ft 3. How much work was required to pump in the fuel?

10 ft 2 ft 10 ft A conical tank if filled to within 2 ft of the top with fuel weighing 57 lb/ft 3. How much work was required to pump in the fuel? 

Example – Pumping Fluids A fuel tank is described below. Determine the work to empty the tank. Assume the pump sits 2 feet above the tank and the fluid weighs 55.6 pounds per cubic foot. Where F(x) is the weight

Where F is weight, and d is distance from pump → (3-y) Foot-pounds Answer