A horizontal spring with K = 110 N/m has one end attached to the wall. A 250g block is pushed onto the free end, compressing the spring by 0.190 . The block is then released, and the spring launches it outward. Neglecting friction, what's its speed when it leaves the spring?
To find the speed of the block when it leaves the spring, we can apply the conservation of mechanical energy.
The potential energy stored in the compressed spring is equal to the kinetic energy of the block when it leaves the spring.
The potential energy stored in the spring is given by the formula:
Potential Energy = 1/2 * k * x^2
Where:
k = spring constant (110 N/m)
x = compression of the spring (0.190 m)
Substituting the given values:
Potential Energy = 1/2 * 110 N/m * (0.190 m)^2
Potential Energy = 1/2 * 110 N/m * 0.0361 m^2
Potential Energy = 0.9905 J
Since the potential energy is converted to kinetic energy when the block leaves the spring, we can set the potential energy equal to the kinetic energy:
0.9905 J = 1/2 * m * v^2
Where:
m = mass of the block (0.250 kg)
v = velocity of the block
Substituting the given values:
0.9905 J = 1/2 * 0.250 kg * v^2
0.9905 J = 0.125 kg * v^2
Dividing both sides by 0.125 kg:
0.9905 J / 0.125 kg = v^2
7.924 J/kg = v^2
Taking the square root of both sides to solve for v:
√(7.924 J/kg) = v
v ≈ 2.81 m/s
Therefore, the speed of the block when it leaves the spring is approximately 2.81 m/s.
To find the speed of the block as it leaves the spring, we can use the principle of conservation of energy. Initially, the system has potential energy stored in the compressed spring. When the block is released, this potential energy is converted into kinetic energy.
The potential energy stored in a compressed spring is given by the equation:
PE = (1/2) * k * x^2
Where:
PE is the potential energy stored in the spring
k is the spring constant (110 N/m)
x is the compression or extension of the spring (0.190 m)
Plugging in the values into the equation:
PE = (1/2) * 110 * (0.190)^2
PE = 2.009 J (Joules)
Since energy is conserved, this potential energy will be converted into kinetic energy when the block leaves the spring.
The kinetic energy (KE) of an object is given by the equation:
KE = (1/2) * m * v^2
Where:
KE is the kinetic energy
m is the mass of the object (250 g = 0.250 kg)
v is the velocity of the object
Equating the potential and kinetic energies:
PE = KE
2.009 J = (1/2) * 0.250 * v^2
Solving for v:
v^2 = 2.009 J / (0.250 kg * 0.5)
v^2 = 16.072 J/kg
Taking the square root:
v = √(16.072 J/kg)
v ≈ 4.01 m/s
Therefore, the speed of the block when it leaves the spring is approximately 4.01 m/s.