Before You Begin

Question #42245

Difficulty: 8/10 | 🟢 Free

The Question

A snowplow moves at a constant speed $v$ directly up a rough inclined plane of angle $\theta$ . The plane is initially covered with a uniform layer of stationary snow of density $\rho$ and cross-sectional area $A$ . The plow collects the snow, which forms a plug of mass $M$ that slides up the incline at speed $v$ just in front of the plow. The coefficient of kinetic friction between the snow plug and the incline is $\mu$ . What is the total force $F$ exerted by the snowplow on the snow plug at the instant its mass is $M$ ?

Options

A.
$\rho A v^2 + M g (\sin\theta + \mu \cos\theta)$
B.
$M g (\sin\theta + \mu \cos\theta)$
C.
$\frac{1}{2}\rho A v^2 + M g (\sin\theta + \mu \cos\theta)$
D.
$\rho A v^2 + M g (\sin\theta - \mu \cos\theta)$

Check Your Answer

Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Variable_Constraint

Develop pure symbolic logic skills by mastering algebraic constraints without relying on numerical data.

Concepts Behind This Question

Equation of Continuity|Power-Velocity Relationship|Friction on Inclined Planes

Why This Question Is Tricky

Common Mistakes

This question tests the application of Newton's Second Law to variable mass systems on an inclined plane.

The lethal trap occurs when students treat the snow plug as a rigid body of constant mass $M$ and apply static equilibrium ( $F = M g \sin\theta + \mu M g \cos\theta$ ), completely ignoring the continuous momentum flux required to accelerate the incoming stationary snow.

Furthermore, it tests the correct sign convention for non-conservative frictional forces and gravitational components opposing the motion.

The distinction between the force required to maintain the velocity of an existing mass and the force required to accelerate new mass ( $v \frac{dm}{dt}$ ) is a classic JEE Advanced discriminator.

Quick Facts About This Question

Question Information

Subject	Physics
Chapter	Mechanical Properties of Fluids - Class 11 Part 2
Concept	Equation of Continuity
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free