Kinetic Energy Problem Set worksheet featuring five word problems for calculating kinetic energy, mass, and velocity in physics.

Kinetic Energy Problem Set

The formula for kinetic energy is given by:
\[
K_E = \frac{1}{2}mv^2
\]
where:
- \( m \) is the mass (in kilograms),
- \( v \) is the speed (in meters per second),
- \( K_E \) is the kinetic energy (in joules).

We will solve each problem step by step.

---

Problem 1:

A 54.43 kg Olympic diver dives off a 10-m platform and reaches a speed of 15.6 m/s. What is her kinetic energy upon impact with the pool water?

#### Given:
- \( m = 54.43 \, \text{kg} \)
- \( v = 15.6 \, \text{m/s} \)

#### Solution:
Using the kinetic energy formula:
\[
K_E = \frac{1}{2}mv^2
\]
Substitute the given values:
\[
K_E = \frac{1}{2} \times 54.43 \times (15.6)^2
\]
First, calculate \( v^2 \):
\[
v^2 = (15.6)^2 = 243.36
\]
Now, substitute back:
\[
K_E = \frac{1}{2} \times 54.43 \times 243.36
\]
Calculate the product:
\[
K_E = 27.215 \times 243.36 = 6630.98 \, \text{J}
\]

#### Final Answer:
\[
\boxed{6630.98 \, \text{J}}
\]

---

Problem 2:

A pick-up truck barrels down a highway with a speed of 27.1 m/s and has a kinetic energy of 326,280 J. What is the mass of the pick-up in kg?

#### Given:
- \( v = 27.1 \, \text{m/s} \)
- \( K_E = 326,280 \, \text{J} \)

#### Solution:
Rearrange the kinetic energy formula to solve for mass \( m \):
\[
K_E = \frac{1}{2}mv^2 \implies m = \frac{2K_E}{v^2}
\]
Substitute the given values:
\[
m = \frac{2 \times 326,280}{(27.1)^2}
\]
First, calculate \( v^2 \):
\[
v^2 = (27.1)^2 = 734.41
\]
Now, substitute back:
\[
m = \frac{2 \times 326,280}{734.41}
\]
Calculate the numerator:
\[
2 \times 326,280 = 652,560
\]
Now, divide:
\[
m = \frac{652,560}{734.41} \approx 888.00 \, \text{kg}
\]

#### Final Answer:
\[
\boxed{888.00 \, \text{kg}}
\]

---

Problem 3:

A standard-sized 0.030 kg NERF bullet hits your opponent with 2.94 J of kinetic energy. At what velocity did the NERF bullet impact your opponent?

#### Given:
- \( m = 0.030 \, \text{kg} \)
- \( K_E = 2.94 \, \text{J} \)

#### Solution:
Rearrange the kinetic energy formula to solve for velocity \( v \):
\[
K_E = \frac{1}{2}mv^2 \implies v^2 = \frac{2K_E}{m} \implies v = \sqrt{\frac{2K_E}{m}}
\]
Substitute the given values:
\[
v = \sqrt{\frac{2 \times 2.94}{0.030}}
\]
First, calculate the numerator:
\[
2 \times 2.94 = 5.88
\]
Now, divide:
\[
\frac{5.88}{0.030} = 196
\]
Finally, take the square root:
\[
v = \sqrt{196} = 14 \, \text{m/s}
\]

#### Final Answer:
\[
\boxed{14 \, \text{m/s}}
\]

---

Problem 4:

If a 113 kg linebacker collides into a tackling dummy with a proper form tackle at a speed of 4.1 m/s, what was the kinetic energy upon impact?

#### Given:
- \( m = 113 \, \text{kg} \)
- \( v = 4.1 \, \text{m/s} \)

#### Solution:
Using the kinetic energy formula:
\[
K_E = \frac{1}{2}mv^2
\]
Substitute the given values:
\[
K_E = \frac{1}{2} \times 113 \times (4.1)^2
\]
First, calculate \( v^2 \):
\[
v^2 = (4.1)^2 = 16.81
\]
Now, substitute back:
\[
K_E = \frac{1}{2} \times 113 \times 16.81
\]
Calculate the product:
\[
K_E = 56.5 \times 16.81 = 956.885 \, \text{J}
\]

#### Final Answer:
\[
\boxed{956.89 \, \text{J}}
\]

---

Problem 5:

An all-star catcher snares a 0.145 kg baseball that collided into the mitt with 117.35 J of kinetic energy. At what velocity did the baseball travel upon impact toward the catcher?

#### Given:
- \( m = 0.145 \, \text{kg} \)
- \( K_E = 117.35 \, \text{J} \)

#### Solution:
Rearrange the kinetic energy formula to solve for velocity \( v \):
\[
K_E = \frac{1}{2}mv^2 \implies v^2 = \frac{2K_E}{m} \implies v = \sqrt{\frac{2K_E}{m}}
\]
Substitute the given values:
\[
v = \sqrt{\frac{2 \times 117.35}{0.145}}
\]
First, calculate the numerator:
\[
2 \times 117.35 = 234.7
\]
Now, divide:
\[
\frac{234.7}{0.145} = 1618.62
\]
Finally, take the square root:
\[
v = \sqrt{1618.62} \approx 40.23 \, \text{m/s}
\]

#### Final Answer:
\[
\boxed{40.23 \, \text{m/s}}
\]

---

Final Answers:

1. \(\boxed{6630.98 \, \text{J}}\)
2. \(\boxed{888.00 \, \text{kg}}\)
3. \(\boxed{14 \, \text{m/s}}\)
4. \(\boxed{956.89 \, \text{J}}\)
5. \(\boxed{40.23 \, \text{m/s}}\)