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MCAT Physical : Work

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Example Question #11 : Work, Energy, And Power

A 1.0kg $\displaystyle 1.0kg$ block slides along a smooth surface at a velocity of $2\frac{m}{s}$ $\displaystyle 2\frac{m}{s}$. The block then encounters a rough surface and continues for six meters before coming to a complete stop. What is the force of friction on the block?

Possible Answers:

$\displaystyle 0N$

$6\textup{N}$ $\displaystyle 6\textup{N}$

$\frac{1}6\textup{N}{}$ $\displaystyle \frac{1}6\textup{N}{}$

$\frac{1}{3}\textup{N}$ $\displaystyle \frac{1}{3}\textup{N}$

$3\textup{N}$ $\displaystyle 3\textup{N}$

Correct answer:

$\frac{1}{3}\textup{N}$ $\displaystyle \frac{1}{3}\textup{N}$

Explanation:

This question can be answered using conservation of energy. Having an understanding of the work-energy theorem allows us the knowledge that changing kinetic energy is a form of work, and that work can be a form of energy. In this particular question, the only force acting on the block to slow it down is the force of friction from the rough surface. As we know, work is equal to force times distance.

W=Fd $\displaystyle W=Fd$

Because the block started with a certain amount of kinetic energy, and then is brought to a complete stop, all of that kinetic energy is transferred to work done by the force of friction over a distance of six meters.

$W=\Delta KE=\frac{1}{2}m(\Delta v)^2$ $\displaystyle W=\Delta KE=\frac{1}{2}m(\Delta v)^2$

Because the 1.0kg $\displaystyle 1.0kg$ block had a velocity of $2\frac{m}{s}$ $\displaystyle 2\frac{m}{s}$ on a smooth surface and ends with a velocity of $0\frac{m}{s}$ $\displaystyle 0\frac{m}{s}$ at rest, we can calculate the change in kinetic energy.

$\Delta KE=\frac{1}{2}(1.0kg)(0\frac{m}{s}-2\frac{m}{s})^2=\frac{1}{2}(1.0kg)(4\frac{m^2}{s^2})$ $\displaystyle \Delta KE=\frac{1}{2}(1.0kg)(0\frac{m}{s}-2\frac{m}{s})^2=\frac{1}{2}(1.0kg)(4\frac{m^2}{s^2})$

$\Delta KE=2J$ $\displaystyle \Delta KE=2J$

The kinetic energy is completely converted to the work done by friction.

$W=\Delta KE=2J$ $\displaystyle W=\Delta KE=2J$

We can then use the first equation for work to determine the force of friction.

$\displaystyle W=Fd=F_f(6m)=2J$

$F_f=\frac{2J}{6m}=\frac{1}{3}N$ $\displaystyle F_f=\frac{2J}{6m}=\frac{1}{3}N$

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Example Question #12 : Work, Energy, And Power

A 2kg box slides down a ramp a distance of two meters before it reaches the ground. The ramp has an angle of 30^o $\displaystyle 30^o$. The coefficient of kinetic friction for the ramp is 0.1.

What is the work done on the box by friction?

Possible Answers:

-4.00J $\displaystyle -4.00J$

-3.24J $\displaystyle -3.24J$

-2.00J $\displaystyle -2.00J$

-1.73J $\displaystyle -1.73J$

-3.39J $\displaystyle -3.39J$

Correct answer:

-3.39J $\displaystyle -3.39J$

Explanation:

Because the force of friction is parallel to the displacement of the box, we can solve for the work done by friction using the following equation.

$W=F\cdot d$ $\displaystyle W=F\cdot d$

Now, we need to define the force of friction or $F_{f}$ $\displaystyle F_{f}$.

$F_f=\mu\cdot m \cdot a$ $\displaystyle F_f=\mu\cdot m \cdot a$

In this formula, $\displaystyle a$ equals acceleration. In this case, acceleration equals acceleration due to gravity in the x-direction; therefore, use the cosine function.

$F_{f}=(0.1)\cdot(2kg)\cdot (g\cdot \cos(30^o))$ $\displaystyle F_{f}=(0.1)\cdot(2kg)\cdot (g\cdot \cos(30^o))$

Insert the value of acceleration due to gravity for $\displaystyle g$.

$F_{f}=0.1\cdot 2kg \cdot -9.8\tfrac{m}{s^2}\cdot \cos(30^o)$ $\displaystyle F_{f}=0.1\cdot 2kg \cdot -9.8\tfrac{m}{s^2}\cdot \cos(30^o)$

$\displaystyle F_f=-1.697N$

Now, we can solve for the work done by friction.

$W_{friction} =F_f\cdot d$ $\displaystyle W_{friction} =F_f\cdot d$

$W_{friction}=(-1.697N)\cdot(2m)$ $\displaystyle W_{friction}=(-1.697N)\cdot(2m)$

$W_{friction} = -3.39J$ $\displaystyle W_{friction} = -3.39J$

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MCAT Physical : Work

Study concepts, example questions & explanations for MCAT Physical

All MCAT Physical Resources

Example Questions

Example Question #11 : Work, Energy, And Power

Example Question #12 : Work, Energy, And Power

All MCAT Physical Resources

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