How Does Friction Affect Block Acceleration on Different Surfaces?

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Friction significantly impacts block acceleration on different surfaces. In the first problem, a 20 kg block on a table and a 10 kg hanging block are connected by a cord, resulting in zero acceleration due to balanced forces, despite the presence of friction. In the second problem, a 40 kg block sliding down a 35° incline experiences a net force that includes gravitational pull and friction, leading to a calculated acceleration. The coefficient of friction alters the net force and thus the acceleration of the blocks. Understanding these dynamics is crucial for solving similar physics problems effectively.
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For the following problems, I don't know how to start . Please explain every step to me with emphasis on the difference in finding the blocks' acceleration accompanied by the coefficient of friction and the case if it was not present. [I have other problems I need to do.]

Find each block's acceleration.

1. A 20 kg block (A) rests on a table; a cord attached to the block extends horizontally to a pulley at the edge of the table. A 10 kg mass, which has the shape of a block, (B) hangs at the end of the cord. The coefficient of [kinetic] friction is 0.5.

2. A 40 kg block slides down an inclined plane. The incline is 35° and the coefficient of friction is 0.2.

Thanks a million. :smile:
 
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1. Draw a force diagram, recall that friction is porportional to the the Normal force and opposes motion. Now sum your forces and solve of acceleration.
2. Essentially the same process as above.

Note: Read the sticky at the top of the homework forum.
 


To find the acceleration of the blocks in these problems, we will use Newton's second law which states that the net force acting on an object is equal to its mass multiplied by its acceleration (F=ma). We will also need to consider the forces acting on the blocks, which include gravity, normal force, and friction.

1. In the first problem, we have two blocks, A and B, connected by a cord. Block A is resting on a table, so the normal force acting on it is equal to its weight, which is mg = (20 kg)(9.8 m/s²) = 196 N. We also have the weight of block B, which is hanging from the cord, pulling down with a force of mg = (10 kg)(9.8 m/s²) = 98 N. Since the blocks are connected by the cord, they will have the same acceleration.

Now, let's consider the horizontal forces acting on block A. The only force acting in that direction is the force of friction, which is equal to the coefficient of friction (0.5) multiplied by the normal force (196 N), giving us a friction force of 98 N. Since the block is not moving horizontally, the friction force must be equal and opposite to the force applied by the cord, which is pulling with a force of 98 N. This means that the net force acting on block A is 0 N, and according to Newton's second law, the acceleration must also be 0 m/s².

Next, let's consider the forces acting on block B. The only force acting on it is the weight pulling down with a force of 98 N. Since the block is moving vertically, the net force acting on it is equal to its mass multiplied by its acceleration, which we will call a. So, 98 N = (10 kg)(a). Solving for a, we get a = 9.8 m/s². This is the acceleration of both blocks, since they are connected by the cord.

2. In the second problem, we have a block sliding down an inclined plane. The forces acting on the block include its weight pulling down the incline with a force of mg = (40 kg)(9.8 m/s²) = 392 N, the normal force acting perpendicular to the incline, and the force of friction acting against the motion of the block. The normal force is equal to the component
 
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