A 0.455 kg wood block is firmly attached to a very light horizontal spring (k = 160 N/m). It is noted that the block-spring system, when compressed 5.0 cm and released, stretches out 2.3 cm beyond the equilibrium position before stopping and turning back. What is the coefficient of kinetic friction between the block and the table
If there is no friction, the block will travel to 5 cm beyond the equilibrium point before turning back.
k=160 N/m
mass, m = 0.455 kg
coefficient of kinetic friction = μ
F=kinetic frictional force
=μm
Potential energy of spring when compressed 5 cm
E1 = (k/2)*(0.05)²
Potential energy of spring when stretched 2.3 cm
E2 = (k/2)*(0.023)²
Lost potential energy
=E1-E2
The energy is lost to overcoming the frictional force over a distance of
D = (5.0 + 2.3) cm
= 0.073 m
Work done by friction
=μ*m*D
Equate energies,
μ*m*D = E1-E2
Solve for μ
The solution was wrong.
E1 is the only energy to consider for the spring, not E2.
E1 = μ*m*D was enough for solving this problem.
Good luck!
E1=mu*mg*D
To determine the coefficient of kinetic friction between the block and the table, we need to consider the forces acting on the block. In this problem, there are two forces that we need to analyze: the force exerted by the spring and the force of kinetic friction.
The force exerted by the spring can be calculated using Hooke's Law, which states that the force exerted by a spring is directly proportional to the displacement from its equilibrium position. The equation for Hooke's Law is:
F = -kx
where F is the force exerted by the spring, k is the spring constant, and x is the displacement from the equilibrium position. Since the block is compressed 5.0 cm before being released, the displacement x is -0.05 m (negative because it is compressed).
Using the given spring constant k = 160 N/m, we can calculate the force exerted by the spring:
F = -kx = -(160 N/m)(-0.05 m) = 8 N
Next, we need to consider the force of kinetic friction. When the block is released and stretches out 2.3 cm beyond the equilibrium position before stopping and turning back, it means that the force of kinetic friction is equal to the force exerted by the spring (8 N) at this point.
The force of kinetic friction can be calculated using the equation:
f_k = μ_k * N
where f_k is the force of kinetic friction, μ_k is the coefficient of kinetic friction, and N is the normal force.
At the point where the block stops and turns back, the normal force is equal to the weight of the block, which can be calculated using the equation:
N = mg
where m is the mass of the block (0.455 kg) and g is the acceleration due to gravity (9.8 m/s²).
N = (0.455 kg)(9.8 m/s²) = 4.469 N
Therefore, the force of kinetic friction is:
f_k = μ_k * N = μ_k * 4.469 N
Since the force of kinetic friction is equal to the force exerted by the spring (8 N), we can set up an equation:
8 N = μ_k * 4.469 N
Solving for μ_k, we get:
μ_k = 8 N / 4.469 N ≈ 1.79
Therefore, the coefficient of kinetic friction between the block and the table is approximately 1.79.