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A golf club strikes a 0.05-kg golf ball in order to launch it from the tee. for simplicity, assume that the average net force applied to the ball acts parallel to the ball's motion, has a magnitude of 6600 n, and is in contact with the ball for a distance of 0.010 mn. determine with what speed does the ball leave the club

a. 49.2 m/s
b. 39.8 m/s
c. 55.9 m/s
d. 53.3 m/s

User MMALSELEK
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1 Answer

3 votes

Final answer:

Using the work-energy principle, we calculate the work done by the force on the golf ball and determine that the ball leaves the club with a velocity of 72.8 m/s, which is not listed among the given options.

Step-by-step explanation:

We will use the work-energy principle to solve this problem. The work done on the golf ball by the golf club is equal to the change in kinetic energy of the golf ball:



Work done (W) = Change in kinetic energy (ΔKE)



Given that the net force (F) is 6600 N and the distance (d) over which this force is applied is 0.010 m (converting from millimeters to meters), we calculate the work done (W) as:



W = F * d = 6600 N * 0.010 m = 66 J



The change in kinetic energy is also the final kinetic energy of the ball (KEf), since it starts from rest. So:



KEf = ½ m * v2



Where m is the mass of the golf ball (0.05 kg) and v is the final velocity. To find v, we can rearrange the equation:



v = sqrt ((2 * KEf) / m)



Substituting the values gives us:



v = sqrt ((2 * 66 J) / 0.05 kg) = 72.8 m/s



Therefore, the correct answer is not listed in the options, as the calculated final velocity is 72.8 m/s.

User Dotz
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