A 2.95 kg block on a horizontal floor is attached to a horizontal spring that is initially compressed 0.0330 m. The spring has force constant 900 N/m. The coefficient of kinetic friction between the floor and the block is 0.42 . The block and spring are released from rest and the block slides along the floor.
What is the speed of the block when it has moved a distance of 0.013 m from its initial position? (At this point, the spring is compressed 0.0200m.)
To find the speed of the block when it has moved a distance of 0.013 m from its initial position, we can break the problem into steps:
Step 1: Calculate the force exerted by the spring.
The force exerted by the spring can be calculated using Hooke's Law: F = -kx, where F is the force, k is the force constant, and x is the displacement from the equilibrium position.
Given:
Force constant, k = 900 N/m
Displacement, x = 0.0200 m (spring compression when the block has moved 0.013 m)
Substituting the values into the formula, we get:
F = -(900 N/m)(0.0200 m) = -18 N
Step 2: Determine the net force acting on the block.
The net force acting on the block can be calculated by considering the force exerted by the spring and the force of kinetic friction.
Given:
Coefficient of kinetic friction, μ = 0.42
Mass of the block, m = 2.95 kg
Acceleration due to gravity, g = 9.8 m/s^2
The force of kinetic friction can be calculated as:
f_k = μ * (mass * gravity)
f_k = 0.42 * (2.95 kg * 9.8 m/s^2) ≈ 12.8414 N
The net force is the vector sum of the spring force and the force of kinetic friction:
Net force, F_net = F + f_k = -18 N + 12.8414 N = -5.1586 N
Step 3: Calculate the acceleration of the block.
Using Newton's second law, F = ma, we can determine the acceleration of the block.
Net force, F_net = mass * acceleration → acceleration = F_net / mass
Substituting the values, we have:
Acceleration, a = (-5.1586 N) / (2.95 kg) ≈ -1.75149 m/s^2 (Note: the negative sign indicates acceleration in the opposite direction of motion)
Step 4: Calculate the final velocity.
Using the kinematic equation, v^2 = u^2 + 2as, where v is the final velocity, u is the initial velocity (which is 0 since the block starts from rest), a is acceleration, and s is the displacement.
Rearranging the equation, we get:
v = sqrt(u^2 + 2as)
Substituting the values, we have:
v = sqrt(0 + 2 * (-1.75149 m/s^2) * 0.013 m) ≈ sqrt(-0.045290899) ≈ 0.213 m/s
Therefore, the speed of the block when it has moved a distance of 0.013 m from its initial position is approximately 0.213 m/s.
To find the speed of the block when it has moved a distance of 0.013 m, we need to analyze the forces acting on the block.
1. Calculate the restoring force of the spring:
The restoring force of the spring, Fs, can be calculated using Hooke's Law:
Fs = -k * x
where k is the force constant of the spring (900 N/m) and x is the displacement from the equilibrium position (initially compressed 0.0330 m, then compressed by an additional 0.0200 m). Thus, x = 0.0330 m + 0.0200 m = 0.0530 m.
Substituting the values, we find:
Fs = -(900 N/m) * (0.0530 m) = -47.7 N
2. Calculate the force of friction:
The force of kinetic friction, Fk, can be calculated using the equation:
Fk = μ * m * g
where μ is the coefficient of kinetic friction (0.42), m is the mass of the block (2.95 kg), and g is the acceleration due to gravity (9.8 m/s^2).
Substituting the values, we find:
Fk = (0.42) * (2.95 kg) * (9.8 m/s^2) = 12.38 N
3. Determine the net force acting on the block:
Since the block is moving to the right, the net force will be the sum of the force of friction and the restoring force of the spring:
Net force = Fk + Fs = 12.38 N - 47.7 N = -35.32 N (negative because it opposes the motion)
4. Apply Newton's second law to find the acceleration:
Newton's second law states that the net force acting on an object is equal to the mass of the object multiplied by its acceleration:
Net force = m * a
Rearranging the equation, we can solve for the acceleration:
a = Net force / m = -35.32 N / 2.95 kg ≈ -11.965 m/s^2
5. Apply the kinematic equation to find the final velocity:
Using the formula:
v^2 = u^2 + 2 * a * s
where v is the final velocity, u is the initial velocity (0 m/s as the block is released from rest), a is the acceleration (-11.965 m/s^2), and s is the distance traveled (0.013 m).
Rearranging the equation and substituting the values, we have:
v = sqrt(u^2 + 2 * a * s) = sqrt(0^2 - 2 * (-11.965 m/s^2) * (0.013 m)) ≈ 0.334 m/s
Therefore, the speed of the block when it has moved a distance of 0.013 m is approximately 0.334 m/s.