Essential Physics

Section 1 review

Energy can take many different forms: mechanical, radiant, nuclear, electrical, chemical, and thermal. Mechanical energy includes kinetic energy and potential energy (both gravitational and elastic). Work and energy are closely related, because energy is the ability to do work, and doing work on an object changes its energy. Potential energy is usually calculated in terms of a reference frame, in which a position of zero energy must be defined. Energy and work are both measured in joules (J). Read the text aloud

Read the text aloud

mechanical energy, work, joule (J), kinetic energy, potential energy, gravitational potential energy, reference frame, elastic potential energy, spring constant

W = F d

E_{k} = \frac{1}{2} m v^{2}

E_{p} = m g h

E_{p} = \frac{1}{2} k x^{2}

Review problems and questions

What forms of energy are found in
1. the Sun?
2. a car’s engine?

A 30 kg boy sitting on a 10 kg go-kart wants to travel down a hill of height 10 m, starting from rest.
1. What is the initial potential energy of the boy and go-kart together?
2. What is their initial kinetic energy?

Answer:

3,920 J
0

Solution:

Asked: initial potential energy E_p,i of the boy and the go-kart
Given: boy and cart combined mass m = 40 kg; hill height h = 10 m; acceleration due to gravity g = 9.8 m/s²
Relationships: E_p = mgh
Solve: $\begin{array}{l} E_{p} & = & m g h \\ = & (40 kg) \times (9.8 m / s^{2}) \times (10 m) = 3, 920 J \end{array}$ Answer: Their initial potential energy is 3,920 J.
Their initial velocity is 0, so their initial kinetic energy is also 0.

A 1,000 kg car traveling 15.0 m/s brakes and comes to a stop after traveling 20.0 m.
1. What is the car’s initial kinetic energy?
2. What is the car’s final kinetic energy?
3. How much work does it take to stop the car?
4. How much constant force is applied in bringing the car to a stop?

Answer:

113 kJ
0
113 kJ
5,630 N

Solution:

Asked: initial kinetic energy E_k,i of the car
Given: car mass m = 1,000 kg; car initial velocity v_i = 15 m/s
Relationships: E_k = ½mv²
Solve: $\begin{array}{l} E_{k} & = & \frac{1}{2} m v^{2} \\ = & \frac{1}{2} (1, 000 kg) \times {(15 m / s)}^{2} = 112, 500 J \end{array}$ Answer: The car’s initial kinetic energy is 112,500 J.
At the end, the car is not moving, so its kinetic energy is zero.
The brakes did enough work to reduce the car’s kinetic energy to zero. Therefore, they did 112,500 J of work.
Asked: force F applied by the brakes
Given: work done W = 112,500 J; work done over distance d = 20 m
Relationships: W = Fd
Solve: Rearrange the equation W = Fd to solve for F: $\begin{array}{l} F & = & \frac{W}{d} \\ = & \frac{112, 500 J}{20 m} = 5, 625 N \end{array}$ Answer: The brakes applied 5,625 N of force.

A baseball outfielder catches a fly ball traveling 25.0 m/s with his gloved hand. When he catches the 0.140 kg baseball, the outfielder’s hand recoils by 10.0 cm. How much force did the outfielder’s hand exert in catching the ball?

Answer: The outfielder exerts 438 N of force on the ball.

Asked: Force F done by the hand on the ball
Given: ball mass m = 0.140 kg; ball velocity v = 25.0 m/s; distance d = 10.0 cm = 0.100 m moved by player’s hand
Relationships: E_k = ½mv², W = Fd
Solve: Calculate the kinetic energy of the ball before being caught:

\begin{array}{l} E_{k} = \frac{1}{2} m v^{2} = \frac{1}{2} (0.140 kg) \times {(25.0 m / s)}^{2} = 43.75 J \\ W = F d \end{array}

Rearrange the equation W = Fd to solve for F:

\begin{array}{l} F & = & \frac{W}{d} \\ = & \frac{43.75 J}{0.100 m} = 437.5 N \end{array}

A horizontal spring with k = 100 N/m is compressed by 10 cm by a 100 g mass.
1. How much elastic potential energy does the compressed spring store?
2. How much elastic potential energy would the compressed spring store if it were compressed the same distance by a 300 g mass?

Answer:

0.5 J
0.5 J

Solution:

Asked: elastic potential energy E_p stored in spring
Given: spring constant k = 100 N/m; compression distance x = 10 cm = 0.1 m
Relationships: E_p = ½kx²
Solve: $E_{p} = \frac{1}{2} k x^{2} = \frac{1}{2} (100 N / m) \times {(0.1 m)}^{2} = 0.5 J$ Answer: The spring stores 0.5 J of potential energy.
Potential energy stored in a spring does not depend on how the spring was compressed. Therefore, in this new case, the stored energy is the same, 0.5 J.


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