A 2kg mass is attached to a spring of spring constant 20N/m. At t=0 the spring is at its relaxed position and the block is moving with a velocity of 4m/s to the right. Assume the positive direction is to the right. What is the speed of the mass when it is 1 meter from the equilibrium position?
A) 1.6 m/s
B) 2.4 m/s
C) 2.5 m/s
D) 3.1 m/s
E) 3.4 m/s
ω=sqrt(k/m) = sqrt(20/2) = 3.16 rad/s
x=Asinωt
v=Aωcosωt
t=0 => cosωt=1
A=v/ω =4/3.16=1.265 m
x=Asinωt
sinωt =x/A=1/1.265 =0.79
cos ωt =sqrt{1-sin²ωt} = 0.612
v₁=Aωcosωt =1.265•3.16•0.612 =2.4464 m
To solve this problem, we need to apply the principle of conservation of mechanical energy. The initial mechanical energy of the system is equal to the final mechanical energy.
The initial mechanical energy is given by the sum of the kinetic energy and the potential energy of the spring:
Initial mechanical energy = (1/2)mv^2 + (1/2)kx^2
The final mechanical energy is the total kinetic energy of the mass when it is 1 meter from the equilibrium position.
Since the mass is attached to a spring, the potential energy of the spring is given by Hooke's law:
Potential energy of the spring = (1/2)kx^2
Applying the conservation of mechanical energy, we have:
(1/2)mv^2 + (1/2)kx^2 = (1/2)kx^2
Simplifying the equation, we have:
(1/2)mv^2 = (1/2)kx^2
Now we can substitute the given values into the equation.
The mass (m) is given as 2 kg.
The spring constant (k) is given as 20 N/m.
The displacement (x) is given as 1 meter.
Substituting these values, we have:
(1/2)(2 kg)v^2 = (1/2)(20 N/m)(1 m)^2
Simplifying further, we have:
v^2 = 10 N
Therefore, the speed of the mass when it is 1 meter from the equilibrium position is the square root of 10 N.
Taking the square root of 10 N, we find that the speed is approximately 3.16 m/s.
Therefore, the correct answer is close to but slightly less than 3.2 m/s.
Among the given options, option D) 3.1 m/s is the closest value, so the correct answer is D) 3.1 m/s.