Two blocks of masses 3kg 6kg as connected by a light inextensible string which passes on a smooth pulley.
Assuming the 3kg masses is sitting on a smooth surface, the bodies will accelerate at?.
Also of the coefficient of the frictional force of the surface is 0.5 the bodies will accelerate.
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3.3 ms-²
To find the acceleration of the two blocks, we can use Newton's second law of motion, which states that the net force on an object is equal to its mass multiplied by its acceleration (F = m*a).
First, let's consider the scenario where there is no friction present on the surface the 3kg block is sitting on. In this case, the only force acting on the 3kg block is its weight (mg), where g is the acceleration due to gravity (9.8 m/s²).
Since the 3kg block is connected to the 6kg block by an inextensible string passing over a smooth pulley, the tension in the string will be the same throughout. Therefore, the tension acting on the 6kg block is also mg.
Now, let's calculate the net force acting on the 6kg block. We have the tension (mg) acting upwards and the weight (mg) acting downwards. The net force (F_net) can be calculated as F_net = mg - mg = 0.
Since the net force on the 6kg block is zero, its acceleration (a) will also be zero. Therefore, the bodies will not accelerate in this scenario.
Now, let's consider the scenario where the coefficient of frictional force (μ) between the 3kg block and the surface is 0.5.
In this case, there is an additional force acting on the 3kg block, which is the frictional force (f_friction). The magnitude of the frictional force can be calculated as f_friction = μ * normal force, where the normal force is equal to the weight of the block (mg).
Therefore, the frictional force will be f_friction = 0.5 * mg.
Now, let's calculate the net force acting on the 6kg block. We have the tension (mg) acting upwards, the weight (mg) acting downwards, and the frictional force (f_friction) acting to the left. The net force (F_net) can be calculated as F_net = mg - mg - f_friction.
Substituting the value of f_friction, we get F_net = mg - mg - 0.5 * mg = -0.5 * mg.
Since the net force on the 6kg block is negative, its acceleration (a) will be in the opposite direction of the net force. Therefore, the bodies will accelerate to the right.
To calculate the magnitude of the acceleration, we can use Newton's second law again:
F_net = m * a
-0.5 * mg = (6kg) * a
Simplifying the equation:
a = (-0.5 * mg) / (6kg)
a = -0.08333g
Therefore, the bodies will accelerate to the right with an acceleration of approximately 0.08333 times the acceleration due to gravity.