a wooden block of mass equal to 15kg is pressed against a spring. The spring has constant of 225N/m. If the spring is compressed 0.15m, how fast will the block travel when released?
To determine how fast the block will travel when released, we can use the principle of conservation of mechanical energy. The potential energy stored in the compressed spring will be converted into the kinetic energy of the block.
1. Calculate the potential energy stored in the compressed spring:
The potential energy stored in a spring (PE_spring) is given by the equation:
PE_spring = 0.5 * k * x^2
where:
k is the spring constant (225 N/m)
x is the compression (0.15 m)
Substituting the given values, we have:
PE_spring = 0.5 * 225 * (0.15)^2
Calculate PE_spring:
PE_spring = 2.53125 J (joules)
2. Since the potential energy stored in the spring will be converted entirely into the kinetic energy of the block, we equate the two energies:
KE_block = PE_spring
where KE_block is the kinetic energy of the block, given by the equation:
KE_block = 0.5 * m * v^2
where:
m is the mass of the block (15 kg)
v is the velocity of the block
Equating the kinetic energy and the potential energy, we have:
0.5 * m * v^2 = PE_spring
3. Solve for v (the velocity of the block):
Rearranging the equation, we get:
v^2 = (2 * PE_spring) / m
Substituting the given values, we have:
v^2 = (2 * 2.53125) / 15
Calculate v:
v^2 = 0.3375
v ≈ √0.3375 ≈ 0.58 m/s
Therefore, when released, the block will travel at a speed of approximately 0.58 m/s.