In the system shown above m1 = 9.5kg and m2 = 2.6kg. If the coefficient of kinetic friction UK = 0.15 and the system is released from rest
No system visible; none posted.
To find the acceleration of the system when it is released from rest, we can use Newton's second law of motion:
ΣF = ma
First, let's analyze the forces acting on the system:
1. Weight force (mg): Each mass experiences a downward weight force due to gravity. The weight force can be calculated using the equation: weight = mass * gravitational acceleration. The gravitational acceleration is approximately 9.8 m/s^2 on the surface of the Earth.
Weight of m1 = m1 * g
Weight of m2 = m2 * g
2. Friction force (f): The friction force opposes the motion of m1. This force depends on the coefficient of kinetic friction (UK) and the normal force. The normal force experienced by m1 is equal to its weight force (m1 * g). The friction force can be calculated using the equation f = UK * normal force.
Friction force on m1 = UK * (m1 * g)
Now, let's apply Newton's second law of motion:
ΣF = ma
The positive direction of motion is defined as downwards. So, the net force acting on the system is the difference between the weight of m1 and the friction force:
Net force = (Weight of m1) - (Friction force on m1)
Net force = (m1 * g) - (UK * (m1 * g))
The net force is also equal to the mass of the system multiplied by its acceleration:
Net force = (m1 + m2) * a
Now we can set up the equation:
(m1 * g) - (UK * (m1 * g)) = (m1 + m2) * a
Rearranging the equation, we can solve for the acceleration:
a = ((m1 * g) - (UK * (m1 * g))) / (m1 + m2)
Now substitute the given values:
m1 = 9.5 kg
m2 = 2.6 kg
UK = 0.15
g = 9.8 m/s^2
a = ((9.5 kg * 9.8 m/s^2) - (0.15 * (9.5 kg * 9.8 m/s^2))) / (9.5 kg + 2.6 kg)
a ≈ 4.18 m/s^2 (rounded to two decimal places)
Therefore, the acceleration of the system when it is released from rest is approximately 4.18 m/s^2.