A 0.41-kg object connected to a light spring with a force constant of 18.6 N/m oscillates on a frictionless horizontal surface. The spring is compressed 4.0 cm and released from rest.
(a) Determine the maximum speed of the object.
(b) Determine the speed of the object when the spring is compressed 1.5 cm
(c) Determine the speed of the object as it passes the point 1.5 cm from the equilibrium position.
(d) For what value of x does the speed equal one-half the maximum speed?
To solve this problem, we can use the principle of conservation of mechanical energy. Since the surface is frictionless, mechanical energy is conserved throughout the motion.
(a) To determine the maximum speed of the object, we need to find the potential energy when the spring is compressed and convert it to kinetic energy when the object reaches maximum speed.
The potential energy stored in the spring when it is compressed 4.0 cm can be calculated using the formula:
PE = (1/2)kx^2
where PE is the potential energy, k is the force constant of the spring, and x is the compression of the spring.
PE = (1/2)(18.6 N/m)(0.04 m)^2 = 0.0372 J
Since the potential energy is converted to kinetic energy at maximum speed, we can equate them:
KE = 0.0372 J
The kinetic energy can be calculated using the formula for kinetic energy:
KE = (1/2)mv^2
where m is the mass of the object and v is the maximum speed.
0.0372 J = (1/2)(0.41 kg)v^2
0.0744 J = (0.205 kg)v^2
v^2 = 0.36 m^2/s^2
v ≈ 0.6 m/s
Therefore, the maximum speed of the object is approximately 0.6 m/s.
(b) To determine the speed of the object when the spring is compressed 1.5 cm, we can repeat the same calculation with a different compression value.
PE = (1/2)(18.6 N/m)(0.015 m)^2 = 0.00209 J
0.00209 J = (1/2)(0.41 kg)v^2
v^2 ≈ 0.01 m^2/s^2
v ≈ 0.1 m/s
Therefore, the speed of the object when the spring is compressed 1.5 cm is approximately 0.1 m/s.
(c) To determine the speed of the object as it passes the point 1.5 cm from the equilibrium position, we can use the conservation of mechanical energy. At this point, the potential energy is converted to kinetic energy.
PE = (1/2)(18.6 N/m)(0.015 m)^2 = 0.00209 J
KE = 0.00209 J
0.00209 J = (1/2)(0.41 kg)v^2
v^2 ≈ 0.01 m^2/s^2
v ≈ 0.1 m/s
Therefore, the speed of the object as it passes the point 1.5 cm from the equilibrium position is approximately 0.1 m/s.
(d) To determine the value of x when the speed equals one-half the maximum speed, we can use the formula for kinetic energy and set it equal to one-half of the maximum kinetic energy.
(1/2)mv^2 = (1/2)(1/2)mv^2
(1/2)(0.41 kg)(0.1 m/s)^2 = (1/2)(1/2)(0.41 kg)(v)^2
0.005205 J = (1/2)(0.205 kg)(v)^2
Solving for v:
(v)^2 = (0.005205 J) / (0.1025 kg)
v ≈ 0.2 m/s
The speed of the object when it's one-half of the maximum speed is approximately 0.2 m/s.
Using the formula for potential energy:
PE = (1/2)kx^2
0.005205 J = (1/2)(18.6 N/m)(x)^2
x^2 ≈ 0.56 m^2
x ≈ ±0.75 m
Therefore, when the speed equals one-half the maximum speed, the value of x is approximately ±0.75 m.