A bowler rolls a 5 kg bowling ball down a magical frictionless bowling alley. The ball accelerates at an average rate of 2 m/s^2. How much force did the bowler apply to the ball?
10 Newtons
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A bowling ball loses momentum of 0.5 kg x m/s when it hit some pins. How much momentum did the pins gain
1. A bowler rolls a 5 kg bowling ball down a magical frictionless bowling alley. The ball accelerates at an average rate of 2 m/s^2. How much force did the bowler apply to the ball?
To determine the force applied by the bowler to the ball, we can use Newton's second law of motion, which states that force (F) is equal to mass (m) multiplied by acceleration (a), or F = m * a.
Given:
Mass of the ball (m): 5 kg
Acceleration of the ball (a): 2 m/s^2
First, we can substitute the known values into the formula:
F = 5 kg * 2 m/s^2
Next, we multiply the mass (5 kg) by the acceleration (2 m/s^2):
F = 10 kg*m/s^2
The unit "kg*m/s^2" is equivalent to a Newton (N), which is the unit of force. Therefore, the force applied by the bowler to the ball is 10 N.