Answer:
0.16 m
Explanation:
Hello
It is possible to solve this problem using the law of conservation of energy, we know that gravitational potential energy and elastic potential energy must be equal
so
[tex]E_{g} =m*g*h\\\\\\E_{s}=\frac{1}{2} k x^{2}\\[/tex]
[tex]E_{g} =m*g*h=\frac{1}{2} k x^{2}[/tex]
Step 1
define and replace
[tex]E_{g} =m*g*h\\\\Let\\[/tex]
m=1.5 kg
h=1.27 m
g=09.81 m/s2
replacing
[tex]E_{g} =m*g*h\\E_{g}=1.5 kg*9.81\frac{m}{s^{2}}*1.27m }\\ E_{g}=25.3098 J[/tex]
Step 2
with Eg Known, now we find x (compression of the spring)
[tex]E_{s}=\frac{1}{2} k x^{2} \\\\x^{2} =\frac{2*E_{s}}{k}\\x=\sqrt{\frac{2*E_{s}}{k}}\\x=\sqrt{\frac{2*25.3095\ J}{ 300 \frac{N}{m} } }\\\\x=0.16 m[/tex]
x=0.16 m
Answer:
0.353 m
Explanation:
According to the law of conservation of energy, the initial energy that the object possesses (in the form of gravitational potential energy) is converted into elastic potential energy of the spring.
The initial gravitational potential energy of the object is:
[tex]E=mgh[/tex]
where
m = 1.50 kg is the mass of the object
g = 9.8 m/s^2 is the acceleration of gravity
h = 1.27 m is the height of the object
The final elastic potential energy of the compressed spring is
[tex]E=\frac{1}{2}kx^2[/tex]
where
k = 300 N/m is the spring constant
x is the compression of the spring
Equalizing the two energies, we find:
[tex]mgh=\frac{1}{2}kx^2\\x=\sqrt{\frac{2mgh}{k}}=\sqrt{\frac{2(1.50 kg)(9.8 m/s^2)(1.27 m)}{300 N/m}}=0.353 m[/tex]
