2Relation of Motion and Work Remember that to change motion, we apply a force for a timeThis causes accelerationRemember the equation vf2 = vi2 + 2ad, which we could re-write as2ad = vf2 – vi2If 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
3Work!The left side of the equation Fd = ½ mvf2 - ½ mvi2 describes something that has been done to cause change in the systemWe call that workW = FdWork 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!
4Work 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!
5More 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!
6Sample Problem 1A 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?
7Sample Problem 1 Write down what you know: Force = 4.5 NDistance = 15 cm = ? mWrite down the formula you'll use:W = FdPut the correct values in the formula:W = (4.5 N)(.15 m)Solve for the unknown:W = NmCheck for units:
8Sample Problem 2What 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?
9Sample problem 2 What you know: Write the equation: Substitute values: Force = 4.5 NDistance = mWrite the equation:W = FdSubstitute values:W = (4.5 N)(0.075 m)Solve:W = NmCheck units!
10Sample Problem 3A 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?
11Sample Problem 3 What you know: Equation: Substitute: Solve: Mass = 7.5 kgForce = (7.5 kg)(9.8 m/s2) (working against gravity)Distance = 8.2 mEquation:W = FdSubstitute:W = (73.5 N)(8.2 m)Solve:W= Nm
12Power Power is the rate of doing work Power is work divided by the time it takes to do the workP = W/tUnits: one joule of power done in one second is 1 Watt
13Sample Problem 4An electric motor lifts an elevator 9 m in 15 s by exerting an upward force of N What is the power produced by the motor?What we know:Distance = 9 mTime = 15 sForce = N
14Sample Problem 4 Equation: Substitute: Solve Check units! P = W/t P = Fd/tSubstitute:P = ( N)(9 m) sSolveP = 7200 WP = 7.2 kWCheck units!
15Remember! 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 JoulesPower units are Watts