A 0.150 block moving vertically upward collides with a light vertical spring and compresses it 4.50 before coming to rest.
If the spring constant is 54.0 , what was the initial speed of the block? (Ignore energy losses to sound and other factors during the collision.)
Duplicate post. Same comment. Units needed.
0.854m/s
To find the initial speed of the block, we can first calculate the potential energy stored in the spring when it is compressed by 4.50 cm, and then use the principle of conservation of energy to find the initial kinetic energy of the block.
The potential energy stored in a spring is given by the formula:
U = (1/2)kx^2
Where U is the potential energy, k is the spring constant, and x is the displacement of the spring from its equilibrium position.
In this case, the spring constant is given as 54.0 N/m and the displacement is 4.50 cm (or 0.045 m). Plugging these values into the formula:
U = (1/2)(54.0 N/m)(0.045 m)^2
U = 0.054 J
The potential energy stored in the spring when it is compressed by 4.50 cm is 0.054 J.
According to the principle of conservation of energy, the initial potential energy of the block is equal to its initial kinetic energy. Therefore, we can equate the potential energy to the kinetic energy:
0.054 J = (1/2)mv^2
Where m is the mass of the block and v is its initial velocity.
In this case, the mass of the block is not given. However, we can use the weight of the block to calculate its mass, as weight is given by the formula:
W = mg
Where W is the weight, m is the mass, and g is the acceleration due to gravity (approximately 9.8 m/s^2).
Assuming the block is near the surface of the Earth, we can use the weight to find the mass:
0.150 kg * 9.8 m/s^2 = 1.47 kg
The mass of the block is 1.47 kg.
Now we can solve the equation for the initial velocity:
0.054 J = (1/2)(1.47 kg)v^2
Simplifying:
0.108 J/kg = v^2
Taking the square root of both sides:
v = sqrt(0.108 J/kg)
Calculating the square root:
v ≈ 0.329 m/s
Therefore, the initial speed of the block is approximately 0.329 m/s.