If the coefficient of kinetic friction is 0.15 find the tension in the string.

In summary, the conversation discusses how to solve a question involving a hanging weight connected to a block on a table with a pulley and a coefficient of kinetic friction. The speaker suggests drawing free body diagrams and using Newton's third law to identify the specific force needed to solve the question.
  • #1
mickeychief
5
0
Can anyone tell me how to solve the below question?

A 9.0kg hanging weight is connected to a string over a pulley to a 5.0 kg block that is sliding on a flat table. If the coefficient of kinetic friction is 0.15 find the tension in the string.
 
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  • #2
Start by drawing free body diagrams for block, hanging mass, and pulley (ideal pulley just changes direction of the tension force in the rope though). Account for all forces...use Newton's third law to help you out. That way you'll be able to identify the specific force you are asked to solve for and how it relates to the others.
 
  • #3


To solve this question, we can use the equation for Newton's Second Law, which states that the net force acting on an object is equal to its mass multiplied by its acceleration. In this case, the net force is the tension in the string.

First, we need to find the acceleration of the 5.0 kg block. We can use the equation for the force of friction, which states that the force of friction is equal to the coefficient of friction multiplied by the normal force. In this case, the normal force is equal to the weight of the block, which is 5.0 kg multiplied by the acceleration due to gravity (9.8 m/s^2). So, the force of friction is 0.15 x 5.0 kg x 9.8 m/s^2 = 7.35 N.

Next, we can set up an equation for the net force on the 5.0 kg block. The net force is equal to the tension in the string minus the force of friction. So, we have:

Net force = Tension - Force of friction

We know that the net force is equal to mass multiplied by acceleration, so we can substitute in the values we know:

5.0 kg x a = Tension - 7.35 N

We also know that the acceleration of the block is the same as the hanging weight, since they are connected by the string. So, we can substitute in the mass of the hanging weight (9.0 kg) for the acceleration:

5.0 kg x 9.8 m/s^2 = Tension - 7.35 N

Solving for tension, we get:

Tension = 49 N + 7.35 N = 56.35 N

Therefore, the tension in the string is 56.35 N.
 

1. What is the coefficient of kinetic friction?

The coefficient of kinetic friction is a measure of the amount of friction between two surfaces in contact when one is in motion. It is represented by the symbol "μ" and is a dimensionless value.

2. How is the coefficient of kinetic friction calculated?

The coefficient of kinetic friction is calculated by dividing the force of friction by the normal force between the two surfaces. It can also be determined experimentally by measuring the force required to keep an object in motion at a constant speed.

3. What does a coefficient of kinetic friction of 0.15 indicate?

A coefficient of kinetic friction of 0.15 means that for every unit of force pushing the two surfaces together, there is a force of 0.15 units resisting the motion between them. This value is relatively low, indicating that the surfaces have a relatively low level of friction when in motion.

4. What is the tension in the string?

The tension in the string refers to the amount of force being applied to the string in order to keep it taut. In this context, the tension is being used to overcome the force of friction and keep the object in motion.

5. How is the tension in the string related to the coefficient of kinetic friction?

The tension in the string is directly related to the coefficient of kinetic friction. The higher the coefficient of kinetic friction, the greater the tension required to overcome it and keep the object in motion. In this case, a coefficient of kinetic friction of 0.15 would require less tension in the string compared to a higher coefficient of friction.

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