A basketball player dunks the ball and momentarily hangs from the rim of the basket. Assume that the player can be considered as a 95.0 kg point...
...mass at a height of 2.0 m 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?
From the information given, we have a player of mass 95 kg hanging from the rim of a basket. The behavior of the basket rim can be simplified to that of a spring with a spring constant of 7.4*10^3 N/m.
The displacement of the basket rim can be determined using Hooke's law. According to Hooke's law, F = -kx, where F is the force causing the displacement, k is the spring constant and x is the displacement from the equilibrium position. The negative sign indicates the tendency of a spring to return to the equilibrium length when displaced.
The player of weight 95 kg exerts a force given by m*g where m is the mass of the player and g is the acceleration due to gravity equal to 9.8 m.s^2. The force is 95*9.8 = 931 N
The displacement due the force x is given by F/k. Here F = 931 N and k = 7.4*10^3 N/m
x = 931/7.4*10^3 = 0.125 m
The displacement of the basket's rim from its horizontal position due to the player hanging from it is 0.125 m.