Work and Power

Relation of Motion and Work Remember that to change motion, we apply a force for a time‏ This causes acceleration Remember the equation vf2 = vi2 + 2ad, which we could re-write as 2ad = vf2 – vi2 If we replace acceleration (a=F/m) and get rid of the 2 and the m on the left by dividing, we get: Fd = ½ mvf2 - ½ mvi2

Work! The left side of the equation Fd = ½ mvf2 - ½ mvi2 describes something that has been done to cause change in the system We call that work W = Fd Work is equal to a constant force exerted on an object in the direction of motion times the distance of the motion. The force MUST be in the direction of the change in motion!

Work it! Whenever work is done, a transfer of energy occurs! Energy can be defined as the ability to do work! Work is done ONLY when the force is being applied to the object!

More about Work! In SI, the units for work are Newton meters, also known as Joules (J)‏ Note that if no change in motion (acceleration) occurs, no WORK is done!

Sample Problem 1 A hockey player exerts a force of 4.5 N over a distance of 15 cm to make a puck slide over the ice. What work has the player done on the puck?

Sample Problem 1 Write down what you know: Force = 4.5 N Distance = 15 cm = ? m Write down the formula you'll use: W = Fd Put the correct values in the formula: W = (4.5 N)(.15 m)‏ Solve for the unknown: W = 0.675 Nm Check for units:

Sample Problem 2 What if the hockey player was only in contact with the puck for half the distance (0.075 m) while using the same force (4.5 N)? What is the work the player has done on the puck?

Sample problem 2 What you know: Write the equation: Substitute values: Force = 4.5 N Distance = 0.075 m Write the equation: W = Fd Substitute values: W = (4.5 N)(0.075 m)‏ Solve: W = 0.3375 Nm Check units!

Sample Problem 3 A rock climber wears a 7.5 kg backpack while climbing a cliff. After 30 minutes, the climber is 8.2 m above the starting point. How much work does the climber do on the backpack? What is the direction of the force? What is the force?

Sample Problem 3 What you know: Equation: Substitute: Solve: Mass = 7.5 kg Force = (7.5 kg)(9.8 m/s2) (working against gravity)‏ Distance = 8.2 m Equation: W = Fd Substitute: W = (73.5 N)(8.2 m)‏ Solve: W= 602.7 Nm

Power Power is the rate of doing work Power is work divided by the time it takes to do the work P = W/t Units: one joule of power done in one second is 1 Watt

Sample Problem 4 An electric motor lifts an elevator 9 m in 15 s by exerting an upward force of 12 000 N What is the power produced by the motor? What we know: Distance = 9 m Time = 15 s Force = 12 000 N

Sample Problem 4 Equation: Substitute: Solve Check units! P = W/t P = Fd/t Substitute: P = (12 000 N)(9 m) 15 s Solve P = 7200 W P = 7.2 kW Check units!

Remember! The FORCE must be in the direction of the work done Write what you know, write the equation, substitute, then solve... in other words SHOW YOUR WORK! Work units are Joules Power units are Watts