Two objects are connected by a light string that passes over a frictionless pulley. The incline is frictionless, m1 = 2.00 kg, m2 = 4.00 kg, and theta= 60.0°.

(b) Find the accelerations of the objects.
Magnitude


(c) Find the tension in the string.
N

(d) Find the speed of each object 3.00 s after being released from rest.
m/s

1 = 4.00 kg and m2 = 9.00 kg are connected by a ...

physics - Objects of masses m1 = 4.00 kg and m2 = 9.00 kg are connected by a ...
physics - Objects of masses m1 = 4.00 kg and m2 = 9

b) 3.58 ms^-2

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To find the accelerations of the objects, we can use Newton's second law of motion. Let's denote the acceleration of m1 as a1 and the acceleration of m2 as a2.

For m1:
1. Find the net force acting on m1. The only force acting on m1 is its weight, which can be broken down into two components: one parallel to the inclined plane (mg*sin(theta)), and one perpendicular to the inclined plane (mg*cos(theta)).
2. The net force acting on m1 is given by the component parallel to the inclined plane: F = m1 * g * sin(theta).
3. Apply Newton's second law: F = m1 * a1. Substitute the net force value to get the equation m1 * g * sin(theta) = m1 * a1.
4. Solve for a1: a1 = g * sin(theta), where g is the acceleration due to gravity (approximately 9.8 m/s^2) and theta is the inclination angle (60.0° in this case).

For m2:
1. Find the net force acting on m2. The only force acting on m2 is its weight, which is directed downwards and has a magnitude of m2 * g.
2. The net force acting on m2 is equal to its weight: F = m2 * g.
3. Apply Newton's second law: F = m2 * a2. Substitute the net force value to get the equation m2 * g = m2 * a2.
4. Solve for a2: a2 = g.

To find the tension in the string, we can use the fact that the tensions in the string on both sides of the pulley are equal. Let's denote the tension in the string as T.

For m1:
1. The weight component parallel to the inclined plane provides the force that causes m1 to move. This force is balanced by the tension in the string, so we have T = m1 * g * sin(theta).

For m2:
1. The weight of m2 provides the force that causes m2 to move downwards. This force is balanced by the tension in the string, so we have m2 * g = T.

To find the speed of each object 3.00 s after being released from rest, we need to use the equations of motion. Since the objects are connected by a string, their speeds will be related based on the pulley's motion. Let's denote the speed of m1 as v1 and the speed of m2 as v2.

Since it is given that the objects are released from rest, their initial velocities are zero. We can use the following kinematic equation for each object:

vf = vi + a * t

For m1:
vi = 0 m/s (since it is released from rest)
a1 = g * sin(theta)
t = 3.00 s

Substituting the values into the equation, we get:
v1 = 0 + (g * sin(theta)) * t

For m2:
vi = 0 m/s (since it is released from rest)
a2 = g
t = 3.00 s

Substituting the values into the equation, we get:
v2 = 0 + g * t

So, the speed of m1 3.00 s after being released is v1 = (g * sin(theta)) * t, and the speed of m2 3.00 s after being released is v2 = g * t.