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Students investigating the motion of golf balls in a medium perform two experiments where golf balls are introduced into different media.  In the first experiment, the balls are dropped vertically downward into a tank of water.  In the second experiment, the balls are compressed against an ideal spring and then propelled horizontally into a tube filled with air, as shown in the figure below.  As the golf ball moves through a medium, it experiences a drag force FD that opposes its motion, as described by the equation:FD=12CDρAv2𝐹D=12𝐶D𝜌𝐴𝑣2where CD is the drag coefficient, ρ is the medium's density, A is the cross-sectional area of the ball, and v is the velocity of the ball. Question 43The students compress a 40 g golf ball a distance of 80 cm against a spring with spring constant k = 100 N/m.  After it is released from the spring, what is the velocity of the ball as it enters the air-filled tube?A.40 m/sB.45 m/sC.56 m/sD.125 m/sSubmit

Question

Students investigating the motion of golf balls in a medium perform two experiments where golf balls are introduced into different media.  In the first experiment, the balls are dropped vertically downward into a tank of water.  In the second experiment, the balls are compressed against an ideal spring and then propelled horizontally into a tube filled with air, as shown in the figure below.  As the golf ball moves through a medium, it experiences a drag force FD that opposes its motion, as described by the equation:FD=12CDρAv2𝐹D=12𝐶D𝜌𝐴𝑣2where CD is the drag coefficient, ρ is the medium's density, A is the cross-sectional area of the ball, and v is the velocity of the ball. Question 43The students compress a 40 g golf ball a distance of 80 cm against a spring with spring constant k = 100 N/m.  After it is released from the spring, what is the velocity of the ball as it enters the air-filled tube?A.40 m/sB.45 m/sC.56 m/sD.125 m/sSubmit

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Solution

To solve this problem, we need to use the principle of conservation of energy. The potential energy stored in the spring when it is compressed is converted into kinetic energy when the ball is released.

The potential energy (PE) stored in a spring is given by the formula:

PE = 1/2 * k * x^2

where k is the spring constant and x is the distance the spring is compressed.

Substituting the given values:

PE = 1/2 * 100 N/m * (0.8 m)^2 = 32 Joules

The kinetic energy (KE) of an object is given by the formula:

KE = 1/2 * m * v^2

where m is the mass of the object and v is its velocity.

We know that the potential energy in the spring is converted into kinetic energy in the ball, so we can set the two equal to each other and solve for v:

32 J = 1/2 * 0.04 kg * v^2

Solving for v gives:

v = sqrt((2 * 32 J) / 0.04 kg) = 40 m/s

So, the velocity of the ball as it enters the air-filled tube is 40 m/s. The answer is A. 40 m/s.

This problem has been solved

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