The Question is from neet-physics laws-of-motion motion-on-a-horizontal-rough-surface, and it was created by the author ABDUL JILANI BASHA

Question

A block of mass $m$ is projected with an initial velocity $v_0$ along a rough horizontal surface with a coefficient of kinetic friction $\mu_k$.  A constant horizontal force $F$, opposite to the direction of motion, also acts on the block. If the block comes to rest after travelling a distance $d$, which of the following expressions correctly relates $v_0$, $\mu_k$, $F$, $m$, $g$, and $d$?

ABDUL JILANI BASHA · Accepted Answer

## CORE INFORMATION
✓ **Answer**: [Option C]
## CONCEPT FRAMEWORK
	**Primary Concept**: Work-Energy Theorem
	**Related Topics**: Kinetic Friction, Work Done by a Force
	**Prerequisites**: Understanding of kinetic energy, work, and friction.
## SOLUTION PATHWAY
**Given Information**:
	* Mass of the block: $m$
	* Initial velocity: $v_0$
	* Coefficient of kinetic friction: $\mu_k$
	* Applied force: $F$ (opposite to motion)
	* Distance travelled: $d$
	* Final velocity: 0

**Method & Steps**:
	1.  Calculate the initial kinetic energy of the block.
		Reasoning: The block starts with kinetic energy due to its initial velocity.
		Formula/Rule: $KE = \frac{1}{2}mv^2$
	2. Calculate the work done by the frictional force. 
		Reasoning: The frictional force opposes motion, doing negative work.
		Formula/Rule: $f_k = \mu_k N$, where N is the normal force. Here, $N = mg$, so $f_k = \mu_k mg$
		Formula/Rule: $W = Fd\cos	heta$, where $	heta$ is the angle between the force and displacement. Here $	heta = 180^\circ$, so $W = -f_k d = -\mu_k mgd$
	3. Calculate the work done by the applied force.
		Reasoning: The applied force also opposes motion, doing negative work.
		Formula/Rule: $W = Fd\cos	heta$. Here $	heta = 180^\circ$, so $W = -Fd$
	4. Apply the work-energy theorem.
		Reasoning: The change in kinetic energy equals the total work done.
		Formula/Rule: $\Delta KE = W_{total}$
	5. Solve for the relationship between the given parameters.
		Working:  \begin{align*}
\Delta KE &= KE_{final} - KE_{initial} \
0 - \frac{1}{2}mv_0^2 &= -\mu_k mgd - Fd \
\frac{1}{2}mv_0^2 &= (\mu_k mg + F)d
\end{align*}
		Key Point: The work done by both friction and the applied force are negative since they oppose the displacement.
	6.  Final Result
		Result: $\frac{1}{2}mv_0^2 = (F + \mu_k mg)d$
		Verification: The derived equation matches option C.
 ## OPTION ANALYSIS
[A] :x: This option incorrectly subtracts the frictional force instead of adding it to the applied force.
[B] :x: This option incorrectly uses $mv_0^2$ instead of $\frac{1}{2}mv_0^2$ and also adds the frictional force incorrectly.
[C] ✓: This option correctly relates initial kinetic energy to the work done by friction and the applied force.
[D] :x: This option incorrectly uses $mv_0^2$ instead of $\frac{1}{2}mv_0^2$ and also subtracts the frictional force instead of adding it.
 ## LEARNING AIDS
	**Common Mistakes**:
		1.[Error 1]: Forgetting to consider the negative work done by friction and the applied force.
			[How to avoid]: Always consider the direction of the force relative to the displacement when calculating work.
		2.[Error 2]: Incorrectly applying the work-energy theorem by not considering the change in kinetic energy.
			[How to avoid]: Always use the formula:  $\Delta KE = KE_{final} - KE_{initial}$
	**Quick Tips**:
		1.[Helpful shortcut]: Remember that the work done by forces opposing motion will be negative.
		2.[Memory aid]:  Work-energy theorem connects the change in kinetic energy to the net work done by all forces.

ABDUL JILANI BASHA · Answer

$\frac{1}{2}mv_0^2 = (F - \mu_k mg)d$

ABDUL JILANI BASHA · Answer

$mv_0^2 = (F + \mu_k mg)d$

ABDUL JILANI BASHA · Answer

The correct Option is "$\frac{1}{2}mv_0^2 = (F + \mu_k mg)d$"

ABDUL JILANI BASHA · Answer

$mv_0^2 = (F - \mu_k mg)d$

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