Key Equations

W = F_{x} \Delta x = Fd \ \cos \circleddash ; Work is equal to the distance an object moves multiplied by the component of the force in the direction that object is moving.

W = work (in Joules; work is just energy being transferred)

Guidance

When an object moves in the direction of an applied force, we say that the force does work on the object. Note that the force may be slowing the object down, speeding it up, maintaining its velocity --- any number of things. In all cases, the net work done is given by this formula:

W = \vec{F}\cdot \vec{d} = \vec{F}\cdot \Delta \vec{x} && \text {Work is the dot product of force and displacement.}

In other words, if an object has traveled a distance  d  under force  \vec{F} , the work done on it will equal to  d  multiplied by the component of \vec{F} along the object's path. Consider the following example of a block moving horizontally with a force applied at some angle:

Here the net work done on the object by the force will be F d \cos \theta.

Example 1

A 1kg ball has been attached to a 2m string and is at rest on a frictionless surface. If you exert a constant force of 10N on the string and pull the ball over the course of 5m and then begin spinning the ball in a circle, after 3 revolutions, what is the total amount of work you have done on the ball?

Solution

Since the centripetal force you exert on the ball in order to make it spin is perpendicular to the ball's path, you do not work on the ball while spinning it in a circle. Therefore, the only work you do on the ball is when you are pulling it in a straight line.

W&=Fd\\W&=10\text{N} * 5\text{m}\\W&=50\text{J}\\

Example 2

A block of mass 5kg is sliding down a ramp inclined at 45 degrees. If the coefficient of kinetic friction between the ramp and the block is 0.3, how much work does the force of friction do as the block slides 3m down to the bottom of the ramp.

Solution

In order to find the work done by friction, we first want to find out the magnitude of the force of friction.

f&=\mu_kN && \text{start with the equation for the force of friction}\\f&=\mu_kmg\cos(30) && \text{substitute the y-component of the weight of the block for the normal force}\\

Now that we have the magnitude of the force of friction, we can plug that into the equation for work.

W&=Fd\\W&=\mu_kmgcos(45)\\W&=0.3*5\text{kg}*9.8\text{m/s}^2*\cos(45)\\W&=10.4J

Watch this Explanation

Time for Practice

  1. You slide down a hill on top of a big ice block as shown in the diagram. Your speed at the top of the hill is zero. The coefficient of kinetic friction on the slide down the hill is zero (\mu_k = 0). The coefficient of kinetic friction on the level part just beneath the hill is 0.1 (\mu_k = 0.1). 
    1. What is your speed just as you reach the bottom of the hill?
    2. How far will you slide before you come to a stop?
  2. Marciel is at rest on his skateboard (total mass 50\;\mathrm{kg}) until he catches a ball traveling with a speed of 50\;\mathrm{m/s}. The baseball has a mass of 2\;\mathrm{kg}. What percent of the original kinetic energy is transferred into heat, sound, deformation of the baseball, and other non-mechanical forms when the collision occurs? 
  3. Investigating a traffic collision, you determine that a fast-moving car (mass 600\;\mathrm{kg}) hit and stuck to a second car (mass 800\;\mathrm{kg}), which was initially at rest. The two cars slid a distance of 30.0\;\mathrm{m} over rough pavement with a coefficient of friction of 0.60 before coming to a halt. What was the speed of the first car? Was the driver above the posted 60\;\mathrm{MPH} speed limit?
  4. Force is applied in the direction of motion to a 15.0\;\mathrm{kg} cart on a frictionless surface. The motion is along a straight line and when t = 0, then x = 0 and v = 0. (The displacement and velocity of the cart are initially zero.) Look at the following graph:
    1. What is the change in momentum during the first 5 sec?
    2. What is the change in velocity during the first 10 sec?
    3. What is the acceleration at 4 sec?
    4. What is the total work done on the cart by the force from 0 - 10 sec?
    5. What is the displacement after 5 sec?
  5. Force is applied in the direction of motion to a 4.00\;\mathrm{kg} cart on a frictionless surface. The motion is along a straight line and when t = 0, v = 0 and x = 0. look at the following graph: 
    1. What is the acceleration of the cart when the displacement is 4\;\mathrm{m}?
    2. What work was done on the cart between x = 3\;\mathrm{m} and x = 8\;\mathrm{m}?
    3. What is the total work done on the cart between 0 -10\;\mathrm{m}?
    4. What is the speed of the cart at x = 10\;\mathrm{m}?
    5. What is the impulse given the cart by the force from 1 - 10\;\mathrm{m}?
    6. What is the speed at x = 8\;\mathrm{m}?
    7. How much time elapsed from when the cart was at x = 8 to when it got to x = 10\;\mathrm{m}?

Answers to Selected Problems

  1. a. 10 \;\mathrm{m/s} b. 52\;\mathrm{m}
  2. 96%
  3. 43.8 \;\mathrm{m/s}

Typical Pressurized Water Reactors (PWR) reactors built in the 1970's produce about 1100 megawatts, whilst the latest designs range up to around 1500 megawatts.  That is 1,500,000,000 Joules/second.

A windmill farm, by comparison, using hundreds of individual windmills produces about 5 megawatts.  That is 5,000,000 Joules/second (assuming the wind is blowing).

Power

Power is defined as the rate at which work is done or the rate at which energy is transformed.

Power=\frac{Work}{Time}

In SI units, power is measured in Joules per second which is given a special name, the watt, W.

1.00 watt = 1.00 J/s

Another unit for power that is fairly common is horsepower.

1.00 horsepower = 746 watts

Example Problem:  A 70.0 kg man runs up a long flight of stairs in 4.0 s.  The vertical height of the stairs is 4.5 m.  Calculate the power output of the man in watts and horsepower.

Solution:  

The force exerted must be equal to the weight of the man = mg = (70.0 \ \text{kg})(9.80 \ \text{m/s}^2) = 686 \ \text{N}

W = Fd = (686 \ \text{N})(4.5 \ \text{m}) = 3090 \ \text{N m} = 3090 \ \text{J}

P=\frac{W}{t}=\frac{3090 \ \text{J}}{4.0 \ \text{s}}=770 \ \text{J/s}=770 \ \text{W}

P = 770 \ \text{W} = 1.03 \ \text{hp}

Since P = \frac{W}{t} and W = Fd, we can use these formulas to derive a formula relating power to the speed of the object that is produced by the power.

P=\frac{W}{t}=\frac{Fd}{t}=F \frac{d}{t}=Fv

The velocity in this formula is the average speed of the object during the time interval.

Example Problem:  Calculate the power required of a 1400 kg car if the car climbs a 10° hill at a steady 80. km/h.

Solution:  80. km/h = 22.2 m/s

In 1.00 s, the car would travel 22.2 m on the road surface but the distance traveled upward would be (22.2 m)(sin 10°) = (22.2 m)(0.174) = 3.86 m.  The force in the direction of the upward motion is the weight of the car = (1400 kg)(9.80 m/s2) = 13720 N.

W = Fd = (13720 \ \text{N})(3.86 \ \text{m}) = 53,000 \ \text{J}

Since this work was done in 1.00 second, the power would be 53,000 W.

If we calculated the upward component of the velocity of the car, we would divide the distance traveled in one second by one second and get an average vertical speed of 3.86 m/s.  So we could use the formula relating power to average speed to calculate power.

P = Fv = (13720 \ \text{N})(3.86 \ \text{m/s}) = 53,000 \ \text{W}

Summary

  • Power is defined as the rate at which work is done or the rate at which energy is transformed.
  • \text{Power}=\frac{\text{Work}}{\text{Time}}
  • \text{Power} = \text{Force} \times \text{velocity}

Last modified: Monday, 4 January 2016, 11:33 AM