An 87-kg box is attached to a spring with a force constant of 82 N/m. The spring is compressed 11 cm and the system is released.
(a)What is the speed of the box when the spring is stretched by 7.0 cm?
(b)What is the maximum speed if the box?
To solve this problem, we can use the principle of conservation of mechanical energy.
(a) To find the speed of the box when the spring is stretched by 7.0 cm, we need to equate the initial potential energy stored in the compressed spring to the kinetic energy of the box when the spring is stretched.
The potential energy stored in a spring is given by the formula: PE = (1/2)kx^2,
where k is the force constant of the spring and x is the displacement from its equilibrium position.
Given:
Mass of the box (m) = 87 kg
Force constant of the spring (k) = 82 N/m
Initial displacement (x1) = -0.11 m (negative because it is compressed)
Final displacement (x2) = -0.07 m (negative because it is stretched)
Initial potential energy (PE1) = (1/2)kx1^2
Final kinetic energy (KE2) = (1/2)mv^2
Since energy is conserved, we can set PE1 equal to KE2:
(1/2)kx1^2 = (1/2)mv^2
Substituting the given values:
(1/2)(82 N/m)(-0.11 m)^2 = (1/2)(87 kg)v^2
Simplifying the equation:
v^2 = [(82 N/m)(0.0121 m^2)] / 87 kg
v^2 = (0.16 N m) / 87 kg
v^2 = 0.0018391
v ≈ √0.0018391
v ≈ 0.0429 m/s
Therefore, the speed of the box when the spring is stretched by 7.0 cm is approximately 0.0429 m/s.
(b) To find the maximum speed of the box, we need to calculate the speed when the spring is at its maximum extension, which means the displacement is 0.
Using the same formula as before, but with x2 = 0:
(1/2)kx2^2 = (1/2)mv^2
Substituting the given values:
(1/2)(82 N/m)(0)^2 = (1/2)(87 kg)v^2
0 = (1/2)(87 kg)v^2
Since the mass is not zero, the maximum speed (v) will be zero.
Therefore, the maximum speed of the box is zero, as the box comes to rest when the spring is at its maximum extension.
To find the answer to the given questions, we can use the principles of energy conservation and Hooke's law.
First, let's find the potential energy stored in the spring when it is compressed by 11 cm. The potential energy stored in the spring is given by the equation:
Potential Energy = (1/2) * k * x^2
Where:
k is the force constant of the spring
x is the displacement of the spring from its equilibrium position
Given:
k = 82 N/m
x = 0.11 m (11 cm converted to meters)
Potential Energy = (1/2) * 82 * (0.11)^2
= 0.5 * 82 * 0.0121
= 0.5 * 0.9942
= 0.4971 J
Now, let's find the kinetic energy of the box when the spring is stretched by 7.0 cm. At maximum elongation, all the potential energy stored in the spring is converted into kinetic energy of the box, according to the principle of conservation of energy:
Potential Energy = Kinetic Energy
Kinetic Energy = (1/2) * m * v^2
Where:
m is the mass of the box
v is the velocity/speed of the box
Given:
m = 87 kg
Potential Energy = 0.4971 J
0.4971 = (1/2) * 87 * v^2
v^2 = (2*0.4971) / 87
v^2 = 0.0114
v = sqrt(0.0114)
v ≈ 0.107 m/s
Therefore, the speed of the box when the spring is stretched by 7.0 cm is approximately 0.107 m/s.
To find the maximum speed of the box, we can use the concept that at maximum elongation, all the potential energy is converted into the kinetic energy of the box. Therefore,
Potential Energy = Kinetic Energy
0.4971 = (1/2) * 87 * v_max^2
v_max^2 = (2*0.4971) / 87
v_max^2 = 0.0114
v_max = sqrt(0.0114)
v_max ≈ 0.107 m/s
Hence, the maximum speed of the box is approximately 0.107 m/s.