
The work–energy theorem states that the net work done on an object equals that object’s change in kinetic energy. Work also can change an object’s gravitational potential energy. Energy changes can be represented by the symbol ΔE, and they can be positive or negative. When acting on moving objects, friction transforms some kinetic energy into thermal energy. Air friction (also called air resistance) can prevent a falling object from converting all of its potential energy into kinetic energy. Efficiency divides the energy one gets out of a system by the energy that one puts in. Efficiency can never be greater than 100%, and often is a great deal less.

work–energy theorem, friction, efficiency


$${W}_{net}=\text{\Delta}{E}_{k}$$

 
 $$\eta =\frac{{E}_{out}}{{E}_{in}}$$



Review problems and questions 

 A 10 kg gokart driven by a 30 kg boy starts from rest and rolls down a hill of height 10 m.
 What is the predicted speed of the boy and gokart at the bottom of the hill?
 The boy’s actual speed at the bottom of the hill is measured to be 10 m/s. Does this agree with your prediction? Why or why not?

Answer:  The speed of the boy and gokart at the bottom of the hill will be 14 m/s.
 Since the measured speed is less than the predicted speed, the boy lost speed and kinetic energy owing to friction.
Solution: Asked: speed v of the boy and gokart
Given: height h = 10 m of hill; acceleration due to gravity g = 9.8 m/s^{2} Relationships: E_{p} = mgh, E_{k} = ½mv^{2}, total energy is conserved Solve: $$mgh=\frac{1}{2}m{v}^{2}$$Rearrange equation to solve for v:$$\begin{array}{lll}v\hfill & =\hfill & \sqrt{2gh}\hfill \\ \hfill & =\hfill & \sqrt{2\left(9.8\text{\hspace{0.17em}}\text{m}/{\text{s}}^{\text{2}}\right)\times \left(10\text{\hspace{0.17em}}\text{m}\right)}=14\text{\hspace{0.17em}}\text{m}/\text{s}\hfill \end{array}$$Answer: The speed of the boy and gokart at the bottom of the hill will be 14 m/s.
 Since the measured speed is less than the predicted speed, the boy lost speed and kinetic energy owing to friction.

 A man applies 500 N of horizontal force to push a 100 kg wooden crate 20 m across a floor.
 What is the work that the man performs upon the crate over the 20 m distance?
 When sliding, the crate experiences a friction force of 200 N in the direction opposite its motion. What is the work that the friction does upon the crate over the 20 m distance?
 What is the net work done on the crate?
 What is the change in the crate’s kinetic energy?
 Calculate the efficiency with which the man’s work is turned into the crate’s kinetic energy.

Answer:  The man performs 10,000 J of work upon the crate.
 The man performs −4,000 J of work upon the crate. (Note that the sign of the work is negative because the force opposes the crate’s motion.)
 The net work on the crate is 6,000 J.
 The crate gains 6,000 J of kinetic energy.
 The man’s work is transformed into crate kinetic energy with an efficiency of 0.6, or 60%.
Solution:
 Asked: work W_{man} done by the man on the crate
Given: applied force F_{man} = 500 N; distance d = 20 m traveled by crate Relationships: W_{man} = F_{man}d Solve: W_{man} = F_{man} × d = (500 N)×(20 m) = 10,000 J Answer: The man performs 10,000 J of work upon the crate.  Asked: work W_{fric} done by friction on the crate
Given: kinetic friction force F_{fric} = −200 N; distance d = 20 m traveled by the crate Relationships: W_{fric} = F_{fric}d Solve: W_{fric} = F_{fric} × d = (−200 N)×(20 m) = −4,000 J Answer: The man performs −4,000 J of work upon the crate. (Note that the sign of the work is negative because the force opposes the crate’s motion.)  Asked: net work W_{net} done on the crate
Given: work W_{man} done by the man; work W_{fric} done by friction Relationships: Net work = Sum of all types of work done on the crate over the time period under study. Solve: W_{net} = W_{man} + W_{fric} = 10,000 J + (−4,000 J) = 6,000 J Answer: The net work on the crate is 6,000 J.  Asked: change in kinetic energy (ΔE_{k}) of the crate
Given: net work W_{net} = 6,000 J done on the crate Relationships: ΔE_{k} = W_{net} Solve: ΔE_{k} = W_{net} = 6,000 J Answer: The crate gains 6,000 J of kinetic energy.  Asked: efficiency η characterizing the work done by the man
Given: total work W_{man} = 10,000 J done by the man; change in kinetic energy ΔE_{k} = 6,000 J for the crate Relationships: η = E_{out}/E_{in} Solve: η = ΔE_{k}/W_{man} = (6,000 J)/(10,000 J) = 0.6 Answer: The man’s work is transformed into crate kinetic energy with an efficiency of 0.6, or 60%. (The other 40% is consumed working against friction.)

 A moving hockey stick hits a stationary 160 g hockey puck (m = 0.16 kg) and applies a horizontal force F to the puck. The two objects remain in contact for the first 0.25 m (25 cm) of the puck’s motion. The puck then slides away at a speed v of 20 m/s. What force F is delivered to the puck?

Answer: The puck is subjected to a force of 128 N (about 29 lb).
Asked: force F applied to the hockey puck Given: puck mass m = 0.16 kg; distance d = 0.25 m through which the force acts; final puck speed v = 20 m/s Relationships: W = Fd; W_{net} = ΔE_{k}; E_{k}= ½mv^{2} Solve: The puck goes from zero kinetic energy when stationary to a final kinetic energy when departing at 20 m/s. Its change in kinetic energy is therefore$$\begin{array}{l}{E}_{f}=\frac{1}{2}(0.16\text{kg}){\left(20\text{m/s}\right)}^{2}=32\text{J}\hfill \\ \text{\Delta}{E}_{k}={E}_{f}{E}_{i}=32\text{J}0\text{J}=32\text{J}\hfill \end{array}$$The puck gains 32 J of kinetic energy. Therefore, the net work done on the puck also is 32 J, since the work–energy theorem states$${W}_{net}=\text{\Delta}{E}_{k}=32\text{J}$$ Neglecting friction, we assume that no horizontal force acts on the puck other than the push from the hockey stick. Therefore, the net work done on the puck is all done by the stick. Work equals the product of force and distance, and the distance is a given. Therefore we can solve for the applied force:$$F=\frac{{W}_{net}}{d}=\frac{32\text{J}}{0.25\text{m}}=128\text{N}$$

 A weightlifter grabs a 50 kg barbell resting on the floor, raises it 2 m, then holds it in place.
 What is the barbell’s kinetic energy before being lifted?
 What is the barbell’s kinetic energy when the lift is complete?
 What is the barbell’s change in kinetic energy?
 What is the net work done on the barbell?
 How much work does the weightlifter perform on the barbell?
 Do your answers to questions d and e match?
 Is the weightlifter’s pull the only force acting upon the barbell while it is being lifted?

 The barbell is resting on the floor, so its kinetic energy is zero.
 Once lifted, the barbell is being held in place. Again, its kinetic energy is zero.
 Since its kinetic energy was zero both before and after the lift, its change in kinetic energy equals zero.
 According to the work–energy theorem, the net work done on the barbell equals zero.
 The work done by the weightlifter is not zero. He applies an average force equal to the barbell’s weight through a distance of 2 m. The barbell weighs 50 kg × 9.8 m/s^{2}, or 490 N. The work done by the weightlifter therefore is 490 N × 2 m, or 980 J.
 The work done by the weightlifter (980 J) does not equal the net work done on the barbell (0 J).
 Gravity (weight) also acts upon the barbell during the lift. As the barbell rises, gravity pulls down, doing negative work on the barbell. This explains how the net work can be zero even though the weightlifter does positive, nonzero work on the barbell.

 What does each hazard represent in the illustration at right?

 gases under pressure (in a gas cylinder)
 explosives: contents may explode, detonate, or otherwise react violently
 health hazard: contents are toxic and/or carcinogenic to humans, or may cause breathing problems

 When reading through an investigation, you realize that it requires you to use a chemical with which you are unfamiliar.
 Where can you find the information to identify whether or not the chemical is hazardous?
 Where can you find the information on what particular classifications of hazards the chemical poses?
 Where can you find information on what protective gear you might have to wear?
 Where can you find instructions for safe handling of the material?
 Where can you find information on whether or not the material, when disposed, is considered hazardous waste?
 Where can you find instructions or a procedure for safe disposal of the substance if it is hazardous?

The answer to all the questions is: on the Safety Data Sheets for the chemical. Most (if not all) of the information can usually also be found on the product’s labeling, particularly if it has been labeled according to the GHS.

 At the front of Nkrumah’s classroom there is a large bottle with hydrochloric acid, but he only needs to use a small amount at his bench. Nkrumah follows good handling procedures to transfer some HCl into a smaller bottle to take to his bench. What else should he do?

He should label the smaller bottle with all the information from the original bottle, such as the name of the substance and its various hazards and precautions.

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