Question

A(n) 79.5 g ball is dropped from a height of 51.3 cm above a spring of negligible mass. The ball compresses the spring to a maximum displacement of 4.57537 cm. The acceleration of gravity is 9.8 m/s 2 . x h Calculate the spring force constant k. Answer in units of N/m.

Answer 1

Answer:

The spring constant is 4159.02 N/m.

Explanation:

Given:

Mass of the ball (m) = 79.5 g = 0.0795 kg [1 g = 0.001 kg]

Height of the ball above spring (h) = 51.3 cm = 0.513 m [1 cm = 0.01 m]

Compression in the spring (x) = 4.57537 cm = 0.0457537 m

Total vertical displacement of the ball is equal to the sum of height above spring and compression of the spring. So,

Total vertical height (h+x) = 51.3 cm + 4.57537 cm = 55.87537 cm = 0.5587537 m

Now, as per energy conservation, the total energy of the ball at any position is always a constant.

So, energy possessed by the ball the highest point is equal to the energy possessed by the ball at the lowest point.

Energy at the highest point is due to gravitational potential energy only and energy at the lowest point is due to elastic potential energy only.

So, GPE = EPE

$mg(h+x)=\frac{1}{2}kx^2$

Here 'k' is the spring constant.

Now, plug in all the values and solve for 'k'. This gives,

$0.795\times 9.8\times 0.5587537=\frac{1}{2}k\times (0.0457537)^2\\\\4.35325\times 2=0.0020934k\\\\k=\frac{8.7065}{0.0020934}=4159.02\ N/m$

Therefore, the spring constant is 4159.02 N/m.