Masses on a pulley- a 150g mass and 100g mass objects on either side of a pulley. the larger mass fell through a distance of 199cm with constant acceleration.
1) find the acceleration and net force on the masses by using the time 1.26 seconds, constant acceleration equations and newton's second law.
2) assuming no friction or air resistance in theroy, calculate the net force and then acceleration of both masses.
To solve these problems, we can use Newton's second law, which states that the net force (F_net) acting on an object is equal to its mass (m) multiplied by its acceleration (a): F_net = m * a.
1) To find the acceleration and net force on the masses using the given time and constant acceleration equations, we need to use the kinematic equation:
d = vt + (0.5)at^2
where d is the distance, v is the initial velocity, a is the acceleration, and t is the time.
Since the larger mass fell through a distance of 199 cm (or 1.99 m) with constant acceleration, the equation becomes:
1.99 = 0*t + (0.5)*a*(1.26)^2
Simplifying the equation gives us:
1.99 = 0.795*a
Dividing both sides by 0.795 gives us the acceleration:
a = 2.506 m/s^2
Now, let's find the net force on the masses:
For the 150g mass (0.15 kg), the net force is:
F_net = m * a = 0.15 * 2.506 = 0.3759 N
For the 100g mass (0.1 kg), the net force is:
F_net = m * a = 0.1 * 2.506 = 0.2506 N
2) Assuming no friction or air resistance, the net force acting on both masses will be the force due to gravity, which is equal to their respective masses multiplied by the acceleration due to gravity (g ≈ 9.8 m/s^2).
Since the masses are on opposite sides of the pulley, the net force acting on both masses would be the difference between their individual forces due to gravity.
For the 150g mass:
F_net = m * g = 0.15 * 9.8 = 1.47 N
For the 100g mass:
F_net = m * g = 0.1 * 9.8 = 0.98 N
Now, we can find the acceleration using Newton's second law:
For the larger mass (0.15 kg):
F_net = m * a
1.47 = 0.15 * a
a = 9.8 m/s^2
For the smaller mass (0.1 kg):
F_net = m * a
0.98 = 0.1 * a
a = 9.8 m/s^2
Hence, in both cases (assuming no friction or air resistance), the net force on each mass is the same, and they both experience an acceleration of 9.8 m/s^2.