A 5 kg box is pushed against a K= 40 n/m spring on a horizontal surface and the spring is compressed 0.12m. The box is released and travels for 1.4 meters before comin to a stop. What is the coefficent of friction between the floor and the box. How fast was the box moving when it lost contact with the spring
I suggest you try an energy conservation approach. The work done against friction will equal the initial system total energy. From that you can deduce the friction force and coefficient.
The kinetic energy when contact with the spring is lost equals the initial stored spring potential energy, less a small amount for friction work while moving 0.12 m
To find the coefficient of friction between the floor and the box, we can use the equation:
Frictional force (Ff) = Normal force (N) * coefficient of friction (μ)
To find the coefficient of friction, we need to calculate the frictional force and the normal force:
1. Calculate the gravitational force acting on the box:
Weight (W) = mass (m) * acceleration due to gravity (g)
= 5 kg * 9.8 m/s^2
= 49 N
2. The normal force acting on the box is equal to the weight of the box (since it is on a horizontal surface and not accelerating vertically):
N = W
= 49 N
3. Calculate the force exerted by the spring using Hooke's Law:
Force exerted by the spring (F) = spring constant (k) * displacement (x)
Given that the spring constant, k, is 40 N/m and the displacement is 0.12 m, we have:
F = 40 N/m * 0.12 m
= 4.8 N
4. The frictional force acting on the box opposes its motion and is equal to the force exerted by the spring:
Ff = F
= 4.8 N
5. Now, we can find the coefficient of friction:
Ff = N * μ
4.8 N = 49 N * μ
μ = 4.8 N / 49 N
≈ 0.098 (rounded to three decimal places)
Therefore, the coefficient of friction between the floor and the box is approximately 0.098.
To find the speed of the box when it loses contact with the spring, we can use conservation of mechanical energy. Assuming there are no other dissipative forces:
1. The initial energy stored in the compressed spring is converted entirely into the kinetic energy of the box and the work done against friction.
Initial potential energy stored in the spring (PEs) = 0.5 * k * x^2, where k is the spring constant and x is the displacement of the spring.
Initial kinetic energy of the box (KEi) = 0, as the box is at rest initially.
Work done against friction (Wf) = Ff * d, where Ff is the frictional force and d is the distance traveled.
2. The final kinetic energy of the box (KEf) is given by 0.5 * m * v^2, where m is the mass of the box and v is the velocity of the box.
Using the conservation of mechanical energy principle:
PEs + KEi - Wf = KEf
As the box comes to a stop, the final kinetic energy (KEf) is 0.
Hence, we have:
0.5 * k * x^2 + 0 - Ff * d = 0.5 * m * v^2
Substituting the given values, we get:
0.5 * 40 N/m * (0.12 m)^2 - 4.8 N * 1.4 m = 0.5 * 5 kg * v^2
Simplifying:
0.288 N - 6.72 N = 2.5 kg * v^2
-6.432 N = 2.5 kg * v^2
v^2 = -6.432 N / 2.5 kg
v^2 ≈ -2.5736 m^2/s^2
Since velocity (v) cannot be negative in this context, the box did not leave the spring. Please check the given values or rephrase the question if necessary.