A 250g wood block firmly attached to a horizontal spring slides along a table with a coefficient of friction of 0.40. A force of 30.N compresses the spring 20cm. If the spring is released from this position what will be the block's speed when it passes equilibrium position?
k = 30 N/.2 m = 150 N/m
PE of spring = (1/2) k x^2 = 75(.04)=3 Joules
normal force = m g
friction force = .4 m g
work done against friction = .2 *.4 * mg
= .08*.25*9.81 = .196 Joules
so Ke left = 3 - .196 = 2.8 Joules
(1/2) m v^2 = 2.8
calculate v
To find the speed of the block when it passes the equilibrium position, we need to consider the conservation of mechanical energy. We know that the initial potential energy of the compressed spring is equal to the final kinetic energy of the block.
1. First, let's find the potential energy of the compressed spring.
- The force applied to compress the spring is 30 N.
- The displacement of the spring is 20 cm = 0.2 m (converted to meters).
- The formula for potential energy stored in a spring is: PE = (1/2)kx^2, where k is the spring constant and x is the displacement.
- Rearranging the formula, we get: k = (2PE) / x^2.
2. Now, let's find the spring constant (k) using the given information.
- The potential energy (PE) of the compressed spring is equal to the work done to compress it.
- The formula for work is: W = F * d * cos(θ), where F is the force applied, d is the displacement, and θ is the angle between the force and displacement.
- In our case, the force (F) is 30 N and the displacement (d) is 0.2 m, and since the force and displacement are in the same direction, the angle between them is 0 degrees.
- Therefore, the work done to compress the spring is W = 30 N * 0.2 m * cos(0°).
- Since cos(0°) = 1, the work done is W = 30 N * 0.2 m = 6 J (Joules).
- Since the potential energy of the compressed spring is equal to the work done, we have PE = 6 J.
Now, substituting the potential energy (PE) and displacement (x) into the formula for the spring constant (k), we get:
k = (2 * 6 J) / (0.2 m)^2 = 60 N/m.
3. With the spring constant determined, we can calculate the speed of the block at the equilibrium position.
- The potential energy at the equilibrium position is zero because the spring is not compressed or stretched.
- Therefore, all the initial potential energy is converted into kinetic energy when the block is at the equilibrium position.
- The formula for kinetic energy is: KE = (1/2)mv^2, where m is the mass of the block and v is its speed.
- Rearranging the formula, we get: v = √(2KE / m).
Since the initial potential energy (PE) is equal to the final kinetic energy (KE), we substitute PE = KE = 6 J, and the mass (m) is given as 250 g = 0.25 kg.
Plugging these values into the formula, we get:
v = √(2 * 6 J / 0.25 kg) ≈ √(48) ≈ 6.93 m/s.
Therefore, the block's speed when it passes the equilibrium position is approximately 6.93 m/s.