Answer:
Work = 9000J
Explanation:
Work done on the spring is 60J
mass of block 60kg
extension on the spring 0.1cm
Workdone = [tex]\frac{1}{2}[/tex]Ke²
where K is the force constant of the spring, Inputting the values
we have 90 = [tex]\frac{1}{2}[/tex]×K×0.1²
K = [tex]\frac{180}{0.01}[/tex]
K = 18000 N/cm
How much work is required to stretch the spring by 1 cm from its equilibrium position, we have
Workdone = [tex]\frac{1}{2}[/tex]Ke²
work = [tex]\frac{1}{2}[/tex]×18000×1²
Work = 9000J
= 9KJ
Answer:
W = 9 kJ
Explanation:
We have:
m: mass of the block = 60 kg
W: work = 90 J
x: distance = 0.1 cm = 1.00x10⁻³ m
To find the work we can use the following equation:
[tex]W = F*x = \frac{1}{2}kx^{2}[/tex]
Where:
F: is the force that the block applies to an ideal spring = -kx
First, we need to find the spring constant:
[tex]k = \frac{2W}{x^{2}} = \frac{2*90 J}{(1.00 \cdot 10^{-3} m)^{2}} = 1.80 \cdot 10^{8} N/m[/tex]
Now, with the spring constant we can find the work required to stretch the spring by 1 cm from its equilibrium position:
[tex]W = \frac{1}{2}kx^{2} = \frac{1}{2}1.80 \cdot 10^{8} N/m*(1.00 \cdot 10^{-2} m)^{2} = 9000 J = 9 kJ[/tex]
Therefore, the work required is 9 kJ.
I hope it helps you!
