Question

A spring is used to launch a coffee mug. The 20cm long spring can be compressed by a maximum of 8cm. The mug has a mass of 350 g. What kind of spring do you need to launch the mug at least 2 stories high (7 m) - as measured from the relaxed spring position - using the maximum compression?

Answer 1

Answer:

7503.13 N/m

Explanation:

Use principle of conservation of energy.

Here, energy stored in the spring due to compression shall be utilized in attaining the potential energy of the mug.

Given that,

Length of the spring = 20 cm = 0.20 m

Compression, x = 8 cm = 0.08 m

mass of the mug, m = 350 g = 0.35 kg

h = 7 m

use the expression for energy balance -

(1/2)*k*x^2 = m*g*h

=> k = (2*m*g*h) / x^2

input the values

k = (2*0.35*9.8*7) / 0.08^2

= 7503.13 N/m

Answer 2

Answer:

The kind of spring needed to launch the 0.35 kg mug to 7 m high from start is 7510.78 N/m.

Explanation:

To solve the question, we note that the

Mass m of the mug = 350 g

Height to which mug is to be launched to = 7 m

Length of spring = 20 cm

Therefore we look for the governing equations of motion as follows

Energy required to raise the mug to 7 m = Gravitational Potential of the mug at 7 m and

Gravitational Potential = m·g·h

Where m = mass = 350 g = 0.35 kg

h = height = 7 m

g = Acceleration due to gravity = 9.81  m/s²

Energy of spring to raise the height of the mug to 7 m = $\frac{1}{2}\times k \times x^2$

Where:

K = Spring constant

x = Length of compression or expansion

Equating the Energy supplied to the energy gained we get

m·g·h = $\frac{1}{2}\times k \times x^2$

Making k the subject of the formula gives

$K = \frac{m \times g \times h}{\frac{1}{2} \times x^2}$ = $K = \frac{0.35 \times 9.81 \times 7}{\frac{1}{2} \times 0.08^2}$ = 7510.78 N/m.