How Much Does a Basketball Player Displace the Rim When Dunking?

[aeromat]

Homework Statement

A basketball player dunks the ball and momentarily hangs from the rm of the basket. Assume that the player can be considered as a 95.0kg point mass at a height of 2.0m above the floor. If the basket rim has a spring constant of 7.4 x 10^3 N/m, by how much does the player displace the rim from the horizontal position?

Homework Equations

Conservation of Energy

The Attempt at a Solution

I set the initial instance to be the bottom, where Eg and Ee would be 0. The final instance to be at the top where he is holding onto the rim. There Eg and Ee are present.

0 = Eg' + Ee'
0 = (m)(g)(h) + 1/2(k)(x)^2
-(m)(g)(h)*2 / k = x^2

Woops, I ran into a negative square root. What am I doing wrong here?

 

[varialectio]

Homework Statement

A basketball player dunks the ball and momentarily hangs from the rm of the basket. Assume that the player can be considered as a 95.0kg point mass at a height of 2.0m above the floor. If the basket rim has a spring constant of 7.4 x 10^3 N/m, by how much does the player displace the rim from the horizontal position?

Homework Equations

Conservation of Energy

The Attempt at a Solution

I set the initial instance to be the bottom, where Eg and Ee would be 0. The final instance to be at the top where he is holding onto the rim. There Eg and Ee are present.

0 = Eg' + Ee'
0 = (m)(g)(h) + 1/2(k)(x)^2
-(m)(g)(h)*2 / k = x^2

Woops, I ran into a negative square root. What am I doing wrong here?

To me, this looks like a static problem of Hooke's Law (spring displacement proportional to load) not one of conservation of energy. What does it matter how high above the floor the ring is?

 

[Andrew Mason]

To me, this looks like a static problem of Hooke's Law (spring displacement proportional to load) not one of conservation of energy. What does it matter how high above the floor the ring is?

The spring has to stretch not only the distance required to balance mg but it has to also stop the player's downward motion. So the approach taken is correct. The negative sign is fixed if one uses g = -9.8 m/sec^2 or if one uses a + sign and says that the energy stored in the spring has to equal the loss of potential energy of the player:

$$mgh = \frac{1}{2}kx^2$$

Having said all that, it is not all that clear how the spring is supposed to work. I think you just have to assume that the relationship between horizontal displacement of the hoop edge and the force is F = -kx. So h = x.

AM

 

[aeromat]

Ok I tried to take two approaches to answer this problem.

[1] 0 = Eg + Ee
0 = (m)(-9.81)(h) + 1/2(k)(x)^2
sqrt[(95)(+9.81)(2)(2) / (7.4*10^3) ] = x
x = 0.7097m
x ≈ 0.71m

[2] F = kx
In this case, I assumed that the gravitational force from the player is the force that is pulling on the spring to stretch it, thus (m)(g)

(95)(9.81) / (7.4*10^3) = x
x = 0.1259m
x ≈ 0.13m

The answer at the back is 0.15m..

 

[rwisz]

I would like to clarify that the concept of dunking on a basketball net is not a scientific topic. However, I will provide a response to the given scenario using the principles of physics.

Firstly, the player's weight of 95.0kg can be considered as a point mass located at a height of 2.0m above the floor. The player's potential energy at the top of the jump can be calculated using the formula Eg = mgh, where m is the mass, g is the acceleration due to gravity (9.8 m/s^2), and h is the height.

Eg = (95.0kg)(9.8 m/s^2)(2.0m) = 1862 J

At the top of the jump, the player is also holding onto the rim, which acts as a spring with a spring constant of 7.4 x 10^3 N/m. This means that the rim will experience a displacement, x, from its original horizontal position.

Using the principle of conservation of energy, we can equate the player's initial potential energy to the final potential energy stored in the spring and the player's final potential energy at the top of the jump.

Eg = Ee + Ee'
1862 J = 1/2(k)(x)^2 + (95.0kg)(9.8 m/s^2)(0m)
1862 J = 1/2(7.4 x 10^3 N/m)(x)^2
x = √(1862 J * 2 / 7.4 x 10^3 N/m) = 0.292 m

Therefore, the player displaces the rim by 0.292 m from its original horizontal position. It is important to note that this calculation assumes ideal conditions and may not accurately reflect the actual displacement in a real-life scenario.