The Question is from NEET guide , and it was created by the author Mr. ANIL KUMAR.G

Question

A block of mass m is pushed up a rough inclined plane of angle $	heta$ and coefficient of kinetic friction $\mu_k$ by a constant horizontal force F.  The block moves a distance d along the incline.  Which expression gives the net work done on the block?

Mr. ANIL KUMAR.G · Accepted Answer

## CORE INFORMATION
✓ **Answer**: B
## CONCEPT FRAMEWORK
	**Primary Concept**: Work-Energy Theorem
	**Related Topics**: Work done by a force, Friction, Inclined planes
	**Prerequisites**: Vectors, Forces, Work, Energy
## SOLUTION PATHWAY
**Given Information**:
	Mass of block: m
	Angle of incline: $	heta$
	Coefficient of kinetic friction: $\mu_k$
	Horizontal force: F
	Displacement along incline: d
**Method & Steps**:
	1. **Resolve forces into components along and perpendicular to the incline:**
		Reasoning: Work is done by the component of the force parallel to the displacement.
		$F_{parallel} = F \cos 	heta$
		$F_{perpendicular} = F \sin 	heta$
		$mg_{parallel} = mg \sin 	heta$
		$mg_{perpendicular} = mg \cos 	heta$
	2. **Calculate the force of friction:**
		Working: The normal force N is equal to the sum of the perpendicular components of F and mg.
		$N = mg \cos 	heta + F \sin 	heta$
		$f_k = \mu_k N = \mu_k (mg \cos 	heta + F \sin 	heta)$
	3. **Calculate the net work done:**
		Result: Net work done is the sum of the work done by each force acting along the incline. Work done by gravity and friction is negative as they oppose the motion.
		$W_{net} = (F \cos 	heta)d - (mg \sin 	heta)d - (\mu_k (mg \cos 	heta + F \sin 	heta))d$
		$W_{net} = Fd(\cos 	heta - \mu_k \sin 	heta) - mgd(\sin 	heta + \mu_k \cos 	heta)$
		Verification: The answer matches option B.
## OPTION ANALYSIS
[A] :x: Incorrect. This only accounts for the work done by the horizontal force and the component of gravity along the incline, ignoring friction.
[B] ✓: Correct. This expression correctly accounts for the work done by the applied force, gravity, and friction.
[C] :x: Incorrect. The sign of the friction term related to the applied force is incorrect.
[D] :x: Incorrect. This expression doesn't account for the displacement d and represents force, not work.
## LEARNING AIDS
	**Common Mistakes**:
		1. Incorrectly resolving forces: Ensure forces are resolved along and perpendicular to the incline.
		2. Forgetting the normal force due to the applied force: The horizontal force contributes to the normal force, affecting friction.
	**Quick Tips**:
		1. Draw a free-body diagram to visualize the forces acting on the block.
		2. Remember that work done is the dot product of force and displacement vectors.

Mr. ANIL KUMAR.G · Answer

Fd cos θ - mgd sin θ

Mr. ANIL KUMAR.G · Answer

The correct Option is "Fd(cos θ - μk sin θ) - mgd(sin θ + μk cos θ)"

Mr. ANIL KUMAR.G · Answer

Fd(cos θ + μk sin θ) - mgd sin θ

Mr. ANIL KUMAR.G · Answer

F cos θ - μk mg cos θ

NEET Questions

Select Subject

Question Tags

Question Tags

Question Tags

Question Tags

Question Tags

Question Tags

Question Tags

Question Tags

Question Tags

NEET Subjects

Our Bots

Company