Question

. A spring system with k = 2500 n?m is compressed for 0.5 m to be used in launching a 2 kg ball and make it roll horizontally on the ground where µ = 0.2 between the ball and the ground. The spring is stretched for 0.1 m when the ball is detached from the spring. For how many meters will the ball roll before coming to a stop?

Answer 1

A spring system with $k = 2500 \, \text{n/m}$ is compressed for $0.5 \, \text{m}$ to be used in launching a $2 \, \text{kg}$ ball and make it roll horizontally on the ground where $\mu = 0.2$ between the ball and the ground. The spring is stretched for $0.1 \, \text{m}$ when the ball is detached from the spring. For how many meters will the ball roll before coming to a stop?

Solution.

$$k = 2500 \frac{N}{m}$$ $$x_1 = 0.5 \, \text{m}$$ $$x_2 = 0.1 \, \text{m}$$ $$m = 2 \, \text{kg}$$ $$\mu = 0.2$$

The kinetic energy of the ball goes on the work of the force of friction

$$E_p = E_k - A_{fr}$$ $$\frac{k x_1^2}{2} = \frac{m v^2}{2} - F_{fr} \cdot S$$

On the other side, elastic force minus the friction force equals mass multiplied by acceleration for Newton's second law:

$$F_{el} - F_{fr} = m a$$ $$N - m g = 0$$ $$F_{fr} = N \cdot \mu = m g \mu$$ $$k x_2 - m g \mu = m a$$

We can express the acceleration, which is equal:

$$a = \frac{k x_2 - m g \mu}{m}$$

And the formula of distance is:

$$S = \frac{v^2}{2a}$$

From which we express the speed

$$v^2 = 2 S a$$

And substituting into equation

$$\frac {k x _ {1} ^ {2}}{2} = \frac {m 2 S a}{2} - F _ {f r} \cdot S$$

Simplify

$$\frac {k x _ {1} ^ {2}}{2} = m S a - m g \mu \cdot S$$ $$\frac {k x _ {1} ^ {2}}{2} = m S \frac {k x _ {2} - m g \mu}{m} - m g \mu \cdot S$$

The distance is

$$S = \frac {k x _ {1} ^ {2}}{2 (2 k x _ {2} - 4 \mu m g)} = \frac {2 5 0 0 \cdot 0 . 2 5}{2 \cdot 2 5 0 0 \cdot 0 . 1 - 4 \cdot 0 . 2 \cdot 2 \cdot 1 0} = 1. 3 m$$

Answer: 1.3m