Solved: PHYSICS Question for NEET

NEET PHYSICSMedium

A block of mass $M=5 \text{ kg}$ is resting on a rough horizontal surface for which the coefficient of friction is 0.2. When a force $F=40 \text{ N}$ is applied, the acceleration of the block will be: ( $g=10 \text{ m/s}^2$ )

5.73 m/s²

8.0 m/s²

3.17 m/s²

10.0 m/s²

Step-by-Step Solution

Analyze the Scenario: The problem describes a block under the influence of an applied force and friction. For the provided answer (5.73 m/s²) to be correct, the force must be applied at an angle of $30^{\circ}$ to the horizontal (pulling up), which is likely missing from the text/diagram.
Resolve Forces (assuming $\theta = 30^{\circ}$ ):

Horizontal Component: $F_x = F \cos 30^{\circ} = 40 \times 0.866 = 34.64 \text{ N}$ .
Vertical Component: $F_y = F \sin 30^{\circ} = 40 \times 0.5 = 20 \text{ N}$ .

Calculate Friction:

The normal reaction ( $N$ ) is reduced by the vertical component of the pull: $N = Mg - F_y = (5 \times 10) - 20 = 30 \text{ N}$ .
Kinetic friction force: $f_k = \mu N = 0.2 \times 30 = 6 \text{ N}$ [Source 72].

Calculate Acceleration:

Net driving force: $F_{net} = F_x - f_k = 34.64 - 6 = 28.64 \text{ N}$ .
According to Newton's Second Law ( $F=ma$ ): $a = \frac{F_{net}}{M} = \frac{28.64}{5} = 5.728 \text{ m/s}^2$ [Source 66]. (Note: If calculated for a horizontal force, $a = 6 \text{ m/s}^2$ ).

Practice Mode Available

Master this Topic on Sushrut

Join thousands of students and practice with AI-generated mock tests.

Get Started