Question

1. you are on roller blades on top of a small hill. your potential energy is equal to 1,000 joules. the last time you checked your mass was 60.0 kilograms.

a. what is your weight in newtons?
b. what is the height of the hill?
c. if you skate down this hill, your potential energy will be converted to kinetic energy. at the bottom of the hill, your kinetic energy will be equal to your potential energy at the top. what will be your speed at the bottom of the hill?

1. answer the following:

a. what is the kinetic energy of a 1-kilogram ball is thrown into the air with an initial velocity of 30 m/sec?
b. how much potential energy does the ball have when it reaches the top of its ascent?
c. how high into the air did the ball travel? **ke=pe

1. what is the potential energy of a 3 kilogram-ball that is on the ground?
2. what is the kinetic energy of a 2,000-kilogram boat moving at 5 m/sec?
3. what is the velocity of a 500-kilogram elevator that has 4,000 joules of energy?
4. what is the mass of an object that creates 33,750 joules of energy by traveling at 30 m/sec?
5. in an experiment, one group of students (group a) measures the speed of a 0.1-kilogram car at 2.5 m/sec at the bottom of a hill. another group of students (group b) measures the speed of the car at 3 m/sec at the bottom of the hill. the cars starting position at the top of the hill is one-meter high.

a. what is the potential energy of the car at the beginning of the experiment before its speed is measured?
b. calculate the kinetic energy of the car for group a using the speed (2.5 m/sec) and mass values above.
c. calculate the kinetic energy of the car for group b using the speed (3.0 m/sec) and mass values above.
d. at the bottom of a hill, the kinetic energy of the cars should be equal to the potential energy of the car at the top of the hill. are the kinetic energy values for groups a and b equal to, less than, or greater than the potential energy value?
e. the energy of an object can be converted to heat due to the friction of the car on the hill. the difference between the potential energy of the car and its kinetic energy at the bottom of the hill equals the energy lost due

Explanation:

Step1: Calculate Group A KE

Use $KE=\frac{1}{2}mv^2$, $m=0.1\ \text{kg}$, $v=2.5\ \text{m/s}$
$\text{KE}_A=\frac{1}{2} \times 0.1 \times (2.5)^2 = 0.3125\ \text{J}$

Step2: Calculate Group B KE

Use $KE=\frac{1}{2}mv^2$, $m=0.1\ \text{kg}$, $v=3\ \text{m/s}$
$\text{KE}_B=\frac{1}{2} \times 0.1 \times (3)^2 = 0.45\ \text{J}$

Step3: Calculate initial potential energy

Use $PE=mgh$, $m=0.1\ \text{kg}$, $g=9.8\ \text{m/s}^2$, $h=1\ \text{m}$
$PE=0.1 \times 9.8 \times 1 = 0.98\ \text{J}$

Step4: Compare KE to PE

Both $\text{KE}_A$ and $\text{KE}_B$ are less than $PE$.

Step5: Explain energy difference

Energy is lost to friction/heat, so KE < PE.

Answer:

a. Kinetic energy for group A: $0.3125\ \text{J}$
b. Kinetic energy for group B: $0.45\ \text{J}$
c. The potential energy of the car at the top of the hill is $0.98\ \text{J}$
d. The kinetic energy values for groups A and B are both less than the potential energy value.
e. The difference occurs because some of the car's potential energy is converted to heat (and other forms of energy) due to friction between the car and the hill, so not all potential energy becomes kinetic energy at the bottom.