Question

name
exponents are used in some areas of science when performing
calculations. in physics, scientists can use a formula to find the
kinetic energy of an object. kinetic energy is energy that an
object holds while it is moving. the kinetic energy of an object is
one-half of the object’s mass times its velocity (speed) squared.
e_k = \frac{1}{2}mv^2
scientists can then determine how the kinetic energy transfers to
another object in a collision or how the kinetic energy transforms
into another type of energy, such as sound or light.
a car that weighs 2,800 pounds is traveling at a velocity of
( 2^4 ) miles per hour. use this information to solve problems 1–5.
do not include units with your answers.

1. write three expressions for the kinetic energy of the car, and then

simplify.

1. the velocity of the car increases to ( 2^2 ) times as fast as the original

velocity. write three expressions for the kinetic energy of the car at
the higher velocity, and then simplify.

1. how does the kinetic energy at the higher velocity compare with

the kinetic energy at the lower velocity?

1. suppose a van weighs ( 2^2 ) times as much as the car and is also

traveling at ( 2^4 ) miles per hour. write three expressions for the
kinetic energy of the van, and then simplify.

1. how does the kinetic energy of the van compare with the original

kinetic energy of the car?

Explanation:

Problem 1

Step1: Identify given values

Mass \( m = 2800 \), velocity \( v = 2^4 \). Kinetic energy formula \( E_k=\frac{1}{2}mv^2 \).

Step2: First expression

Substitute \( m = 2800 \), \( v = 2^4 \) into formula: \( \frac{1}{2} \times 2800 \times (2^4)^2 \)

Step3: Simplify first expression

Simplify \( (2^4)^2 = 2^{4\times2}=2^8 \), \( \frac{1}{2} \times 2800 = 1400 \), so \( 1400\times2^8 \)

Step4: Second expression

Rewrite \( 2800 = 28 \times 100 \), so \( \frac{1}{2} \times 28 \times 100 \times (2^4)^2 \)

Step5: Simplify second expression

\( \frac{1}{2} \times 28 = 14 \), \( 14\times100 = 1400 \), \( (2^4)^2 = 2^8 \), so \( 1400\times2^8 \) (same as first simplified)

Step6: Third expression

Use exponent rules: \( \frac{1}{2} \times 2800 \times 2^{8} \) (since \( (2^4)^2 = 2^8 \))

Step7: Simplify third expression

\( \frac{1}{2} \times 2800 = 1400 \), so \( 1400\times2^8 = 1400\times256 = 358400 \)

Step1: New velocity

Original velocity \( v = 2^4 \), new velocity \( v' = 2^2 \times 2^4 = 2^{4 + 2}=2^6 \) (using \( a^m \times a^n = a^{m + n} \))

Step2: First expression

Substitute \( m = 2800 \), \( v' = 2^6 \) into \( E_k=\frac{1}{2}mv^2 \): \( \frac{1}{2} \times 2800 \times (2^6)^2 \)

Step3: Simplify first expression

\( (2^6)^2 = 2^{12} \), \( \frac{1}{2} \times 2800 = 1400 \), so \( 1400\times2^{12} \)

Step4: Second expression

Rewrite \( 2800 = 28 \times 100 \), so \( \frac{1}{2} \times 28 \times 100 \times (2^6)^2 \)

Step5: Simplify second expression

\( \frac{1}{2} \times 28 = 14 \), \( 14\times100 = 1400 \), \( (2^6)^2 = 2^{12} \), so \( 1400\times2^{12} \)

Step6: Third expression

\( \frac{1}{2} \times 2800 \times 2^{12} \)

Step7: Simplify third expression

\( 1400\times2^{12}=1400\times4096 = 5734400 \)

Step1: Find ratio of kinetic energies

Let \( E_{k1} \) be original KE (from Problem 1: \( 358400 \)), \( E_{k2} \) be new KE (from Problem 2: \( 5734400 \))

Step2: Calculate ratio

\( \frac{E_{k2}}{E_{k1}}=\frac{5734400}{358400}=16 = 2^4=(2^2)^2 \)

Step3: Interpret ratio

Since velocity increased by a factor of \( 2^2 \), and KE is proportional to \( v^2 \), \( (2^2)^2 = 2^4 = 16 \). So new KE is 16 times original KE.

Answer:

Expressions: \( \frac{1}{2} \times 2800 \times (2^4)^2 \), \( \frac{1}{2} \times 28 \times 100 \times (2^4)^2 \), \( \frac{1}{2} \times 2800 \times 2^8 \); Simplified: \( 358400 \)

Problem 2