A 2 kg block situated on a rough incline surface (θ= 37o) and is connected to a spring with k = 100 N/m. The block initially at rest. The block moves 20 cm down the incline before stops. Find the coefficient of friction between the block and the surface.
(Can you give all the solution.TQ)
51+2
To find the coefficient of friction between the block and the surface, we can break down the forces acting on the block.
1. Gravitational Force (Fg):
The force due to gravity acting on the block can be calculated using the formula Fg = m * g, where m is the mass of the block (2 kg) and g is the acceleration due to gravity (9.8 m/s²). So, Fg = 2 kg * 9.8 m/s² = 19.6 N.
2. Normal Force (Fn):
The normal force is the force exerted perpendicular to the incline surface. Here, since the block is on an incline, the normal force can be calculated using the formula Fn = m * g * cos(θ), where θ is the angle of the incline (37 degrees). So, Fn = 2 kg * 9.8 m/s² * cos(37 degrees) = 15.59 N.
3. Spring Force (Fs):
The spring force is given by Hooke's Law, which states that the force exerted by a spring is proportional to its displacement. The formula is Fs = k * x, where k is the spring constant (100 N/m) and x is the displacement of the block (20 cm or 0.2 m). So, Fs = 100 N/m * 0.2 m = 20 N.
Now, let's analyze the forces parallel to the incline:
4. Frictional Force (Ff):
The frictional force can be calculated using the equation Ff = μ * Fn, where μ is the coefficient of friction. Since the block is moving upwards, the frictional force will act in the opposite direction to its motion. Therefore, Ff = -μ * Fn.
Next, we can consider the forces perpendicular to the incline:
5. Force Due to Gravity Parallel to the Incline (Fgx):
The force due to gravity acting parallel to the incline can be calculated using the formula Fgx = m * g * sin(θ), where θ is the angle of the incline. So, Fgx = 2 kg * 9.8 m/s² * sin(37 degrees) = 11.67 N. This force acts downhill.
Now, using Newton's second law (ΣF = ma), we can analyze the forces and motion of the block:
ΣF = ma
Fgx - Ff - Fs = ma
Since the block stops, its acceleration is zero. Thus, we have:
Fgx - Ff - Fs = 0
Substituting the known values:
11.67 N - μFn - 20 N = 0
Substituting the value of Fn:
11.67 N - μ * 15.59 N - 20 N = 0
Now, we can solve for μ:
11.67 N - 15.59 μ - 20 N = 0
-15.59 μ = -11.67 N + 20 N
-15.59 μ = 8.33 N
Dividing by -15.59:
μ ≈ -8.33 N / -15.59N
μ ≈ 0.533
Therefore, the coefficient of friction between the block and the surface is approximately 0.533.