Question

A 25 lb bowling ball starts at rest. Henry decides to give the 25 lb bowling ball a push and causes it to accelerate at a rate of 8.9 m/s^2. How much force did Henry apply to the bowling ball?

Answer 1

Henry applied approximately 100.93 newtons of force to a 25 lb bowling ball to accelerate it at a rate of 8.9 m/s², after converting the ball's weight to kilograms using the conversion factor of 1 lb to 0.453592 kg.

Step-by-step explanation:

To determine the force that Henry applied to the 25 lb bowling ball, we can use Newton's second law of motion, which states that the force applied to an object is equal to the mass of the object multiplied by its acceleration (F = ma). First, we need to convert the weight of the bowling ball from pounds to kilograms, knowing that 1 lb is approximately 0.453592 kg. The bowling ball weighs 25 lb, which is equivalent to 25 lb × 0.453592 kg/lb = 11.34 kg.

With the mass of the bowling ball (m) and the acceleration (a) given as 8.9 m/s2, we can calculate the force (F) as follows:

F = m × a = 11.34 kg × 8.9 m/s2 = 100.9256 N

Therefore, Henry applied a force of approximately 100.93 newtons to the bowling ball.