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:
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 = [tex]\frac{1}{2}\times k \times x^2[/tex]
Where:
K = Spring constant
x = Length of compression or expansion
Equating the Energy supplied to the energy gained we get
m·g·h = [tex]\frac{1}{2}\times k \times x^2[/tex]
Making k the subject of the formula gives
[tex]K = \frac{m \times g \times h}{\frac{1}{2} \times x^2}[/tex] = [tex]K = \frac{0.35 \times 9.81 \times 7}{\frac{1}{2} \times 0.08^2}[/tex] = 7510.78 N/m.
