How Does Increasing Mass Affect Acceleration in a Frictional System?

[Gjky424]

In the figure below, assume that the pulleys are massless and frictionless.

https://www.physicsforums.com/attachment.php?attachmentid=4001&stc=1 23.png

The masses of the blocks are Ma=1.50 kg, Mb=3.00 kg, Mc=5.00 kg, and there is friction between the horizontal plane and Mc, (μk≠0). Mc is observed to travel at a constant velocity.

A. Calculate the coefficient of kinetic friction between Mc and the horizontal surface.

B. Mass Mb is now increased by 0.40 kg. Calculate the magnitude of the acceleration of Mc.

 

[Dr.Brain]

Draw the free body diagram taking into account the friction forces, tensions in ropes , mg downwards and Normal Reaction on Mc. As Mc travels without acceleration, equate all forces.

BJ

 

[thepatientmental]

A. To calculate the coefficient of kinetic friction, we can use the formula μk = Ff/N, where Ff is the force of friction and N is the normal force. In this case, the normal force is equal to the weight of block Mc, which is given by N = Mcg, where g is the acceleration due to gravity (9.8 m/s^2). The force of friction can be calculated using the formula Ff = μkN. Since we are given that Mc is moving at a constant velocity, we know that the net force on it is zero. Therefore, the force of friction must be equal in magnitude to the force pulling it down the incline, which is given by Mbgsinθ, where θ is the angle of the incline. Putting all of this together, we get the equation μkMcg = Mbgsinθ. Plugging in the given values, we get μk = (Mbgsinθ)/(Mcg) = (3.00 kg)(9.8 m/s^2)sin(30°)/(5.00 kg)(9.8 m/s^2) = 0.294. Therefore, the coefficient of kinetic friction between Mc and the horizontal surface is 0.294.

B. When Mb is increased by 0.40 kg, the net force on Mc will change. The new net force will be equal to Mbgsinθ + Magsinθ, where Ma is the mass of block A. This is because the tension in the string connecting blocks A and B will also contribute to the net force on Mc. Using Newton's second law (F=ma), we can set this net force equal to the mass of Mc (5.00 kg) times its acceleration. This gives us the equation (Mbgsinθ + Magsinθ) = Mac. Plugging in the given values, we get (3.40 kg)(9.8 m/s^2)sin(30°) = (5.00 kg)a. Solving for a, we get a = 2.71 m/s^2. Therefore, the magnitude of the acceleration of Mc is 2.71 m/s^2.