Answer:
0.045 J
Explanation:
From the question,
The elastic potential energy stored in a spring is given as,
E = 1/2ke²...................... Equation 1
Where E = elastic potential energy, k = spring constant, e = compression.
Given: k = 100 N/m, e = 0.05-0.02 = 0.03 m
Substitute these values into equation 1
E = 1/2(100)(0.03²)
E = 50(9×10⁻⁴)
E = 0.045 J
Hence the right option is 0.045 J
Something super duper uper stuper luper nuper tuper zuper yuper fuper guper huper kuper juper wuper special
Explanation:
where is the question
I did not understood this question
Answer:
a) = 10.22 rad/s
b) = 0.35 m
Explanation:
Given
Mass of the particle, m = 1.1 kg
Force constant of the spring, k = 115 N/m
Distance at which the mass is released, d = 0.35 m
According to the differential equation of s Simple Harmonic Motion,
ω² = k / m, where
ω = angular frequency in rad/s
k = force constant in N/m
m = mass in kg
So,
ω² = 115 / 1.1
ω² = 104.55
ω = √104.55
ω = 10.22 rad/s
If y(0) = -0.35 m and we want our A to be positive, then suffice to say,
The value of coefficient A in meters is 0.35 m
