JEE Main 2025 Apr 3 Shift 2, Physics Q19: Work Energy Theorem And Its Applications

JEE Main2025Apr 3, Shift 2Physics

A block of mass $1$ kg, moving along $x$ with speed $v_{i} = 10$ m/s enters a rough region ranging from $x = 0.1$ m to $x = 1.9$ m. The retarding force acting on the block in this range is $F_{r} = - k x$ N, with $k = 10$ N/m. Then the final speed of the block as it crosses rough region is

A
$10$ m/s
B
$4$ m/s
C
$6$ m/s
D
$8$ m/s

Solution

Apply the work-energy theorem from $x = 0.1$ m to $x = 1.9$ m. Work done by the retarding force:

$W = \int_{0.1}^{1.9} (- k x) d x = - \frac{k}{2} [x^{2}]_{0.1}^{1.9} = - \frac{10}{2} (1. 9^{2} - 0. 1^{2}) = - 5 \cdot (3.61 - 0.01) = - 18$ J.

Energy balance with $m = 1$ kg:

$\frac{1}{2} v_{f}^{2} = \frac{1}{2} (10)^{2} - 18 = 50 - 18 = 32 \Rightarrow v_{f}^{2} = 64$ .

$∴ v_{f} = 8$ m/s.

Concept behind this question

Work Energy Theorem And Its ApplicationsNotes, formulas and examples →

Practice more Physics

Concept-wise practice with instant solutions on Fundamenthol.

Start practicing →

JEE Main 2025 Apr 3 Shift 2, Physics Q19: Work Energy Theorem And Its Applications

Concept behind this question

More previous year questions on Work Energy Theorem And Its Applications

Practice more Physics