A mass (10kg) slides down to the left on an inclined 30-degree surface with a coefficient of friction (Ukinetic=0.2). It is connected over a pulley to a hanging 5 kg mass by a massless stretch-free cord. What is the acceleration up the plane of the mass on the plane? (work needed) The answer: 1.13
To find the acceleration of the mass on the inclined plane, we can use Newton's second law of motion, which states that the net force acting on an object is equal to the mass of the object multiplied by its acceleration (F=ma).
First, let's consider the forces acting on the mass on the inclined plane. There are three forces at play: the gravitational force (mg), the normal force (N), and the frictional force (f).
The gravitational force acting on the mass can be broken down into two components: one parallel to the inclined plane (mg*sinθ) and one perpendicular to the inclined plane (mg*cosθ), where θ is the angle of inclination.
The normal force (N) acts perpendicular to the inclined plane and counterbalances the perpendicular component of the gravitational force (mg*cosθ).
The frictional force (f) acts parallel to the inclined plane and opposes the motion of the mass. Its magnitude is given by the equation f = μN, where μ is the coefficient of friction and N is the normal force.
Since the mass is connected to a hanging mass over a pulley, the tension (T) in the cord is the same on both sides. Therefore, the force due to the hanging mass (5kg) is T.
Now, let's analyze the forces acting on the system:
- For the mass on the inclined plane:
- Gravitational force parallel to the inclined plane (mg*sinθ)
- Frictional force opposite to the motion (-f)
- For the hanging mass:
- Tension force (T)
Applying Newton's second law to each object:
- For the mass on the inclined plane:
ma = mg*sinθ - f
- For the hanging mass:
T = mg
Since T = mg, we can substitute this into the equation for the mass on the inclined plane:
ma = T*sinθ - f
Substituting the expression for f (frictional force) using the coefficient of friction:
ma = T*sinθ - μN
Substituting N with its corresponding component of the gravitational force (mg*cosθ):
ma = T*sinθ - μ(mg*cosθ)
Now, let's rewrite the equation in terms of acceleration (a):
a = (T*sinθ - μ(mg*cosθ))/m
Substituting T = mg:
a = (mg*sinθ - μ(mg*cosθ))/m
Now, let's plug in the given values. We have:
m = 10kg (mass on the inclined plane)
μ = 0.2 (coefficient of friction)
θ = 30 degrees
a = (10*9.8*sin(30) - 0.2*(10*9.8*cos(30)))/10
Evaluating this expression, we find that the acceleration (a) is approximately 1.13 m/s².
Therefore, the acceleration up the inclined plane for the given system is 1.13 m/s².