Respuesta :
Answer:
The spring constant of the spring is 11.65 N/m
Explanation:
It is given that, a 50 g mass hanger hangs motionless from a partially stretched spring. When a 65 gram mass is added to the hanger, the spring stretch increases by 10 cm.
Initial mass, m = 50 g = 0.005 kg
Final mass, m' = 50 g + 65 g = 115 g = 0.115 kg
Initially, the elongation is x (say). The spring stretch increases by 10 cm or 0.1 m.
We know that,
F = k x
Where
k is the spring constant
For the initial condition :
[tex]mg=kx[/tex].............(1)
For the final condition :
[tex](m+m')g=k(x+10)[/tex]........(2)
Solving equation (1) and (2)
[tex]\dfrac{mg}{(m+m')g}=\dfrac{x}{x+0.1}[/tex]
[tex]\dfrac{x}{x+0.1}=\dfrac{0.005}{(0.005+0.115)}[/tex]
x = 0.0042 m
Now put the value of x in equation (1)
[tex]k=\dfrac{mg}{x}[/tex]
[tex]k=\dfrac{0.005\times 9.79}{0.0042}[/tex]
[tex]k=11.65\ N/m[/tex]
Hence, this is the required solution.
The spring constant of the spring will be k=11.65 N/m
What is spring constant?
The spring constant is defined as it is the ratio of force applied on the spring and the deflection of the spring.
It is given that, a 50 g mass hanger hangs motionless from a partially stretched spring. When a 65 gram mass is added to the hanger, the spring stretch increases by 10 cm.
Initial mass, m = 50 g = 0.005 kg
Final mass, m' = 50 g + 65 g = 115 g = 0.115 kg
Initially, the elongation is x (say). The spring stretch increases by 10 cm or 0.1 m.
We know that,
F = k x
Where
k is the spring constant
For the initial condition :
[tex]mg =kx[/tex].............(1)
For the final condition :
[tex](m+m')g=k(x+10)[/tex]........(2)
Solving equation (1) and (2)
[tex]\dfrac{mg}{(m+m')}=\dfrac{x}{x+0.1}[/tex]
x=0.0042 m
Now put the value of x in equation (1)
[tex]k=\dfrac{mg}{x}[/tex]
[tex]k=\dfrac{0.005\times 9.79}{0.0042}[/tex]
[tex]k= 11.65\ \frac{N}{m^2}[/tex]
Hence the spring constant of the spring will be k=11.65 N/m
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