The Question is from neet-physics laws-of-motion , and it was created by the author NAGA SAI

Question

A block is projected up a rough inclined plane with an initial velocity $u$. The coefficient of kinetic friction between the block and the plane is $\mu_k$. The block travels a distance $s$ up the plane before coming to rest.  What is the angle of inclination $	heta$ of the plane?

NAGA SAI · Accepted Answer

## CORE INFORMATION
✓ **Answer**: D
## CONCEPT FRAMEWORK
	**Primary Concept**: Work-Energy Theorem
	**Related Topics**: Inclined Planes, Friction, Kinematics
	**Prerequisites**: Understanding of kinetic energy, work done by forces, and friction.
## SOLUTION PATHWAY
**Given Information**:
	Initial velocity = $u$
	Final velocity = $0$
	Distance traveled = $s$
	Coefficient of kinetic friction = $\mu_k$
	Angle of inclination = $	heta$
**Method & Steps**:
	1.	Calculate the net work done on the block.
		Reasoning: The net work done is the sum of the work done by gravity and friction. The work-energy theorem states that the net work done on an object is equal to the change in its kinetic energy.
		Formula/Rule: $W_{net} = \Delta KE$
	2.	Calculate the work done by gravity.
		Working: The work done by gravity is given by  $W_g = -mgs\sin	heta$ (negative since gravity opposes the motion).
		Key Point: The component of gravity acting along the incline is $mg\sin	heta$.
	3.	Calculate the work done by friction.
		Working: The normal force is $N = mg\cos	heta$, and the frictional force is $f_k = \mu_k N = \mu_k mg\cos	heta$. The work done by friction is $W_f = -f_k s = -\mu_k mgs\cos	heta$ (negative since friction opposes the motion).
		Key Point: The frictional force always opposes the motion.
	4.	Apply the work-energy theorem.
		Working: The initial kinetic energy is $\frac{1}{2}mu^2$, and the final kinetic energy is $0$. So, the change in kinetic energy is $\Delta KE = 0 - \frac{1}{2}mu^2 = -\frac{1}{2}mu^2$.  Therefore, $W_g + W_f = \Delta KE$, which gives $-mgs\sin	heta - \mu_k mgs\cos	heta = -\frac{1}{2}mu^2$.
	5.	Solve for $\sin	heta + \mu_k\cos	heta$.
		Result: Dividing both sides by $-mgs$, we get $\sin	heta + \mu_k\cos	heta = \frac{u^2}{2gs}$.
		Verification: The units are consistent on both sides.
 ## OPTION ANALYSIS
[A] :x: Incorrect, the correct relation is $\sin	heta + \mu_k\cos	heta = \frac{u^2}{2gs}$.
[B] :x: Incorrect, the correct relation is $\sin	heta + \mu_k\cos	heta = \frac{u^2}{2gs}$ and the sign is wrong.
[C] :x: Incorrect, the correct relation is $\sin	heta + \mu_k\cos	heta = \frac{u^2}{2gs}$.
[D] ✓: Correct, the derived relation matches this option.
 ## LEARNING AIDS
	**Common Mistakes**:
		1.[Error 1]: Forgetting to consider the work done by friction.
			[How to avoid]: Always remember to include all forces that do work on the object.
		2.[Error 2]: Incorrectly calculating the component of gravity or the frictional force.
			[How to avoid]: Draw a free-body diagram and carefully resolve the forces into their components.
	**Quick Tips**:
		1.[Helpful shortcut]: Work-energy theorem is a powerful tool to solve problems involving work and energy changes.
		2.[Memory aid]: Remember that the work done by friction is always negative because it opposes the motion.

NAGA SAI · Answer

$\sin	heta + \mu_k\cos	heta = \frac{2gs}{u^2}$

NAGA SAI · Answer

$\sin	heta - \mu_k\cos	heta = \frac{u^2}{2gs}$

NAGA SAI · Answer

$\sin	heta + \mu_k\cos	heta = \frac{u^2}{gs}$

NAGA SAI · Answer

The correct Option is "$\sin	heta + \mu_k\cos	heta = \frac{u^2}{2gs}$"

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