Essential Physics

Section 2 review

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. Read the text aloud

Read the text aloud

work–energy theorem, friction, efficiency

W_{n e t} = Δ E_{k}

Δ E = 0

Δ E = Q + W

η = \frac{E_{o u t}}{E_{i n}}

Review problems and questions

A 30 kg boy sitting on a 10 kg go-kart starts from rest down a hill of height 10 m.
1. What do you predict will be the combined speed of the boy and go-kart when they reach the bottom of the hill?
2. 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 go-kart 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 go-kart
Given: height h = 10 m of hill; acceleration due to gravity g = 9.8 m/s²
Relationships: E_p = mgh, E_k = ½mv², total energy is conserved
Solve: $m g h = \frac{1}{2} m v^{2}$ Rearrange equation to solve for v: $\begin{array}{l} v & = & \sqrt{2 g h} \\ = & \sqrt{2 (9.8 m / s^{2}) \times (10 m)} = 14 m / s \end{array}$ Answer: The speed of the boy and go-kart 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.
1. What is the work that the man performs upon the crate over the 20 m distance?
2. 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?
3. What is the net work done on the crate?
4. What is the change in the crate’s kinetic energy?
5. 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_mand
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_fricd
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 makes contact with a stationary 160 g hockey puck (m = 0.16 kg). The stick 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= ½mv²
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 kg) {(20 m/s)}^{2} = 32 J \\ Δ E_{k} = E_{f} - E_{i} = 32 J - 0 J = 32 J \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_{n e t} = Δ E_{k} = 32 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_{n e t}}{d} = \frac{32 J}{0.25 m} = 128 N

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

Safety hazards

What does each hazard represent in the illustration at right?

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

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?


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