Jayden holds a 3kg block pressed up against a spring on a horizontal and frictionless surface. The spring has a spring constant of 300Nm. The spring is fixed in place, but the block is free to move. The spring is initially compressed by .5 m and the block is at rest. Jayden lets go of the block and it begins to move. What is the speed of the block when the spring is compressed by .4 m? (The block stays in contact with the spring at all times.)
If needed, g=10ms2.
To find the speed of the block when the spring is compressed by 0.4 m, we can use the principle of conservation of mechanical energy. The total mechanical energy is the sum of kinetic energy and potential energy.
1. Initial potential energy (U₁):
U₁ = (1/2) * k * x₁²
where k is the spring constant and x₁ is the initial compression of the spring.
In this case, k = 300 N/m and x₁ = 0.5 m.
U₁ = (1/2) * 300 N/m * (0.5 m)²
= 75 J
2. Final potential energy (U₂):
U₂ = (1/2) * k * x₂²
where x₂ is the final compression of the spring.
In this case, x₂ = 0.4 m.
U₂ = (1/2) * 300 N/m * (0.4 m)²
= 48 J
3. Initial kinetic energy (Kᵢ):
Kᵢ = 0 (since the block is at rest initially)
4. Final kinetic energy (K_f):
K_f = K
Since the total mechanical energy is conserved, we have:
U₁ + Kᵢ = U₂ + K_f
5. Solve for final kinetic energy:
K_f = U₁ - U₂
K_f = 75 J - 48 J
= 27 J
6. Calculate the speed of the block:
K = (1/2) * m * v²
where m is the mass of the block (3 kg) and v is the speed of the block.
Rearrange the equation:
v² = 2K / m
Substituting the values:
v² = 2 * 27 J / 3 kg
= 18 J / kg
v = √(18 J / kg)
= √(18) m/s
= 4.24 m/s (approximately)
Therefore, the speed of the block when the spring is compressed by 0.4 m is approximately 4.24 m/s.