a box whose mass is 0.3 kg is pushed against a spring(k=40n/m) on a horizontal surface and the spring is compressed 0.12m. The box is released and travels for 1.4 meters before coming to a stop. What is the coefficient of friction between the floor and the box? How fast was the box moving when it lost contact with the spring?
To find the coefficient of friction between the floor and the box, we need to calculate the force exerted by the spring and the force of friction acting on the box.
First, let's find the force exerted by the spring using Hooke's Law:
Force exerted by the spring (F_spring) = k * x
where k is the spring constant and x is the displacement of the spring from its equilibrium position.
Given:
Spring constant (k) = 40 N/m
Displacement of the spring (x) = 0.12 m
F_spring = 40 N/m * 0.12 m
F_spring = 4.8 N
Next, we need to calculate the force of friction acting on the box. When the box is moving, the force of friction is equal to the force required to bring it to a stop. This force can be calculated using Newton's second law:
Force of friction (F_friction) = mass * acceleration
Given:
Mass of the box (m) = 0.3 kg
Acceleration of the box (a) = 0 m/s^2 (since it comes to rest)
F_friction = 0.3 kg * 0 m/s^2
F_friction = 0 N
Since the box comes to rest, the force of friction is equal to zero.
The force exerted by the spring is equal to the force of friction, so we can set them equal to each other:
F_spring = F_friction
4.8 N = 0 N
Since the force of friction is zero, there is no friction between the floor and the box. Therefore, the coefficient of friction between the floor and the box is zero.
Now, let's calculate the speed of the box when it loses contact with the spring.
The initial potential energy stored in the spring is equal to the final kinetic energy of the box:
Potential Energy (PE) = Kinetic Energy (KE)
The potential energy of the spring is given by:
Potential Energy (PE) = 1/2 * k * x^2
Given:
Spring constant (k) = 40 N/m
Displacement of the spring (x) = 0.12 m
PE = 1/2 * 40 N/m * (0.12 m)^2
PE = 0.288 J
The kinetic energy of the box is given by:
Kinetic Energy (KE) = 1/2 * mass * velocity^2
Given:
Mass of the box (m) = 0.3 kg
Velocity of the box (v) = unknown
KE = 1/2 * 0.3 kg * v^2
Now, setting the two energies equal:
0.288 J = 1/2 * 0.3 kg * v^2
Simplifying the equation:
0.576 J = 0.15 kg * v^2
v^2 = 0.576 J / 0.15 kg
v^2 = 3.84 m^2/s^2
Taking the square root of both sides:
v = sqrt(3.84 m^2/s^2)
v ≈ 1.96 m/s
Therefore, the box was moving at approximately 1.96 m/s when it lost contact with the spring.