(a) 392 N/m
Hook's law states that:
[tex]F=k\Delta x[/tex] (1)
where
F is the force exerted on the spring
k is the spring constant
[tex]\Delta x[/tex] is the stretching/compression of the spring
In this problem:
- The force exerted on the spring is equal to the weight of the block attached to the spring:
[tex]F=mg=(2.0 kg)(9.8 m/s^2)=19.6 N[/tex]
- The stretching of the spring is
[tex]\Delta x=15 cm-10 cm=5 cm=0.05 m[/tex]
Solving eq.(1) for k, we find the spring constant:
[tex]k=\frac{F}{\Delta x}=\frac{19.6 N}{0.05 m}=392 N/m[/tex]
(b) 17.5 cm
If a block of m = 3.0 kg is attached to the spring, the new force applied is
[tex]F=mg=(3.0 kg)(9.8 m/s^2)=29.4 N[/tex]
And so, the stretch of the spring is
[tex]\Delta x=\frac{F}{k}=\frac{29.4 N}{392 N/m}=0.075 m=7.5 cm[/tex]
And since the initial lenght of the spring is
[tex]x_0 = 10 cm[/tex]
The final length will be
[tex]x_f = x_0 +\Delta x=10 cm+7.5 cm=17.5 cm[/tex]
(a) The spring constant of the spring is 392 N/m
(b) Length of the spring is 17.5 cm
[tex]\texttt{ }[/tex]
Hooke's Law states that the length of a spring is directly proportional to the force acting on the spring.
[tex]\boxed {F = k \times \Delta x}[/tex]
F = Force ( N )
k = Spring Constant ( N/m )
Δx = Extension ( m )
[tex]\texttt{ }[/tex]
The formula for finding Young's Modulus is as follows:
[tex]\boxed {E = \frac{F / A}{\Delta x / x_o}}[/tex]
E = Young's Modulus ( N/m² )
F = Force ( N )
A = Cross-Sectional Area ( m² )
Δx = Extension ( m )
x = Initial Length ( m )
Let us now tackle the problem !
[tex]\texttt{ }[/tex]
Given:
initial length of spring = Lo = 10 cm
mass of object = m = 2.0 kg
extension of the spring = x = 15 - 10 = 5 cm = 0.05 m
mass of second object = m' = 3.0 kg
Asked:
a. spring constant of the spring = k = ?
b. length of spring = L = ?
Solution:
[tex]F = kx[/tex]
[tex]mg = kx[/tex]
[tex]k = mg \div x[/tex]
[tex]k = 2.0 ( 9.8 ) \div 0.05[/tex]
[tex]\boxed {k = 392 \texttt{ N/m}}[/tex]
[tex]\texttt{ }[/tex]
[tex]F' = kx'[/tex]
[tex]m' g = k x'[/tex]
[tex]x' = ( m' g ) \div k[/tex]
[tex]x' = ( 3.0 (9.8) ) \div 392[/tex]
[tex]x' = 0.075 \texttt{ m} = 7.5 \texttt{ cm}[/tex]
[tex]\texttt{ }[/tex]
[tex]L = Lo + x'[/tex]
[tex]L = 10 + 7.5[/tex]
[tex]\boxed {L = 17.5 \texttt{ cm}}[/tex]
[tex]\texttt{ }[/tex]
[tex]\texttt{ }[/tex]
Grade: College
Subject: Physics
Chapter: Elasticity
