A 3 kg block collides with a massless spring of spring constant 90 N/m attached to a wall. The speed of the block was observed to be 1.5 m/s at the moment of collision. The acceleration of gravity is 9.8 m/s^2. How far does the spring compress if the surface on which the mass moves is frictionless?
Answer in units of cm.
.23 meters
To find the distance the spring compresses, we need to use the principle of conservation of mechanical energy. We know that the initial mechanical energy of the block is equal to its final mechanical energy when the spring is fully compressed.
The initial mechanical energy consists of the kinetic energy of the block, given by (1/2)mv^2, where m is the mass of the block (3 kg) and v is its velocity (1.5 m/s).
The final mechanical energy consists of the potential energy stored in the compressed spring, given by (1/2)kx^2, where k is the spring constant (90 N/m) and x is the distance the spring compresses.
Since the surface is frictionless, we can neglect any energy losses due to friction.
Equating the initial and final mechanical energies, we get:
(1/2)mv^2 = (1/2)kx^2
Substituting the given values, we have:
(1/2)(3 kg)(1.5 m/s)^2 = (1/2)(90 N/m)x^2
Simplifying the equation:
2.25 J = 45 N/m * x^2
Dividing both sides by 45 N/m:
0.05 J = x^2
Taking the square root of both sides:
x = sqrt(0.05 J)
Converting the answer to centimeters:
x = sqrt(0.05 J) * 100 cm/m
Calculating:
x ≈ 3.16 cm
Therefore, the spring compresses approximately 3.16 cm.