Question

how much more kinetic energy does a 6-kilogram bowling ball have when it is rolling at 16 mph (7.1 meters per second) than when it is rolling at 14 mph (6.2 meters per second)?
$ke = \frac{1}{2}mv^2$
(1 point)
\bigcirc 151.2 j
\bigcirc 35.9 j
\bigcirc 266.5 j
\bigcirc 1.3 j

Explanation:

Step1: Calculate KE at 7.1 m/s

Use \( KE = \frac{1}{2}mv^2 \), \( m = 6 \) kg, \( v = 7.1 \) m/s.
\( KE_1 = \frac{1}{2} \times 6 \times (7.1)^2 = 3 \times 50.41 = 151.23 \) J.

Step2: Calculate KE at 6.2 m/s

Use \( KE = \frac{1}{2}mv^2 \), \( m = 6 \) kg, \( v = 6.2 \) m/s.
\( KE_2 = \frac{1}{2} \times 6 \times (6.2)^2 = 3 \times 38.44 = 115.32 \) J.

Step3: Find the difference

Subtract \( KE_2 \) from \( KE_1 \): \( 151.23 - 115.32 = 35.91 \approx 35.9 \) J.

Answer:

35.9 J