Final answer:
The spring constant (k) of the spring compressed by a 7.05 kg bowling ball dropped from 1.69 m and compressing the spring by 0.422 m is approximately 1795.7 N/m.
Step-by-step explanation:
To determine the spring constant (k) of the spring compressed by a 7.05 kg bowling ball dropped from 1.69 m, we need to use the conservation of energy principle. The potential energy lost by the bowling ball when dropped is converted into the elastic potential energy stored in the spring when compressed. The gravitational potential energy (PE) of the bowling ball before it hits the spring can be calculated using the formula PE = mgh, where m is the mass (7.05 kg), g is the gravitational acceleration (9.81 m/s2), and h is the height (1.69 m). The elastic potential energy (EPE) of the compressed spring is given by EPE = (1/2)kx2, where k is the spring constant and x is the compression distance (0.422 m). Setting the gravitational potential energy equal to the elastic potential energy allows us to solve for k:
mgh = (1/2)kx2
k = 2mgh/x2
k = 2(7.05 kg)(9.81 m/s2)(1.69 m)/(0.422 m)2
After calculation, the spring constant (k) is found to be approximately:
k = 1795.7 N/m
The spring constant of the large spring must therefore be approximately 1795.7 N/m to compress by 0.422 m under the weight of a 7.05 kg bowling ball dropped from 1.69 m.