A golf club strikes a 0.045-kg golf ball in order to launch it from the tee. Forsimplicity, assume that the average net force applied to the ball acts parallel to theball’s motion, has a magnitude of 6800 N, and is in contact with the ball for adistance of 0.010 m. With what speed does the ball leave the club?
Question
A golf club strikes a 0.045-kg golf ball in order to launch it from the tee. Forsimplicity, assume that the average net force applied to the ball acts parallel to theball’s motion, has a magnitude of 6800 N, and is in contact with the ball for adistance of 0.010 m. With what speed does the ball leave the club?
Solution
To solve this problem, we can use the work-energy theorem, which states that the work done on an object is equal to the change in its kinetic energy.
Step 1: Calculate the work done on the golf ball. Work done (W) = Force (F) x Distance (d) W = 6800 N * 0.010 m = 68 Joules
Step 2: Set the work done equal to the change in kinetic energy. The initial kinetic energy of the golf ball is 0 (since it starts from rest), so the work done on the ball is equal to its final kinetic energy.
Step 3: Solve for the final speed. The kinetic energy (KE) of an object is given by the formula KE = 1/2 * mass (m) * velocity^2 (v^2). Setting this equal to the work done gives: 68 Joules = 1/2 * 0.045 kg * v^2
Step 4: Solve the equation for v (velocity). First, multiply both sides of the equation by 2: 136 = 0.045 * v^2 Then, divide both sides by 0.045: v^2 = 136 / 0.045 = 3022.22 Finally, take the square root of both sides to solve for v: v = sqrt(3022.22) = 54.97 m/s
So, the golf ball leaves the club with a speed of approximately 54.97 m/s.
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