What is the total work done on the block by the force of friction as the block moves a distance L up the incline? Express the work done by friction in terms of any or all of the va... What is the total work done on the block by the force of friction as the block moves a distance L up the incline? Express the work done by friction in terms of any or all of the variables μ, m, g, θ, L, and F1.
Understand the Problem
The question is asking for the total work done on a block by the force of friction as it moves up an inclined plane a distance L. It seeks an expression for the work done by friction in terms of various physical variables such as the coefficient of friction, mass, gravitational acceleration, angle of incline, distance moved, and applied force.
Answer
$$ W_{fric} = -\mu m g L \cos(\theta) $$
Answer for screen readers
$$ W_{fric} = -\mu m g L \cos(\theta) $$
Steps to Solve
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Identify the force of friction The force of friction ($F_{fric}$) can be calculated using the formula: $$ F_{fric} = \mu F_{N} $$ where $F_{N}$ is the normal force.
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Determine the normal force On an incline, the normal force ($F_{N}$) can be expressed as: $$ F_{N} = mg \cos(\theta) $$ where $m$ is the mass of the block, $g$ is the gravitational acceleration, and $\theta$ is the angle of the incline.
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Calculate the friction force Substituting the expression for the normal force into the friction force equation gives: $$ F_{fric} = \mu (mg \cos(\theta)) $$
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Define the work done by friction The work done by friction ($W_{fric}$) is given by: $$ W_{fric} = -F_{fric} \times d $$ where $d = L$ (the distance moved up the incline).
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Substitute the expression for frictional force into the work formula Insert $F_{fric}$ from step 3 into the work formula: $$ W_{fric} = -(\mu (mg \cos(\theta))) L $$
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Final expression for work done by friction Thus, the expression for the total work done by the friction force as the block moves a distance $L$ up the incline is: $$ W_{fric} = -\mu m g L \cos(\theta) $$
$$ W_{fric} = -\mu m g L \cos(\theta) $$
More Information
This expression shows that the work done by friction is negative, indicating that it opposes the direction of the block's movement up the incline. The more friction, the more negative work done, and hence the more energy required to pull the block up.
Tips
- Confusing the normal force: Remember that it changes on an incline and is not simply $mg$.
- Neglecting the direction of friction: Always consider that friction acts opposite to the direction of motion, leading to negative work.
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