Question

at is the kinetic energy of a 0.135 kg baseball thrown at 40.0 m/s? KE = .5*mv2
a
87.0 J
b
108 J
c
216 J
d
54.0 J

Answer 1

To find the kinetic energy (KE) of the baseball, we can use the formula:

\[
KE = \frac{1}{2}mv^2
\]

where:
- \( m \) is the mass of the object (in kilograms),
- \( v \) is the velocity of the object (in meters per second).

Given:
- \( m = 0.135 \) kg,
- \( v = 40.0 \) m/s.

Now we can plug in the values:

\[
KE = \frac{1}{2} \times 0.135 \, \text{kg} \times (40.0 \, \text{m/s})^2
\]

Calculating \( (40.0 \, \text{m/s})^2 \):

\[
(40.0 \, \text{m/s})^2 = 1600 \, \text{m}^2/\text{s}^2
\]

Now substitute back into the kinetic energy formula:

\[
KE = \frac{1}{2} \times 0.135 \, \text{kg} \times 1600 \, \text{m}^2/\text{s}^2
\]

\[
KE = 0.0675 \times 1600
\]

\[
KE = 108 \, \text{J}
\]

So the kinetic energy of the baseball is:

**b) 108 J**