A block of mass 5 kg is placed on a rough inclined surface as shown in the figure. If $\overrightarrow{F}_{1}$ is the force required to just move the block up the inclined plane and $\overrightarrow{F}_{2}$ is the force required to just prevent the block from sliding down, then the value of $|\overrightarrow{F}_{1}|-|\overrightarrow{F}_{2}|$ is: [Use $g=10$ m/s$^{2}$]

JEE Main · 2024 · 31 Jan 2024 · Shift 2 · Laws of Motion

A block of mass 5 kg is placed on a rough inclined surface as shown in the figure.
If

F_{1}

is the force required to just move the block up the inclined plane and

F_{2}

is the force required to just prevent the block from sliding down, then the value of

∣ F_{1} ∣ - ∣ F_{2} ∣

is: [Use

g = 10

m/s

^{2}

]

From the figure, the incline angle is

30°

and coefficient of kinetic friction

μ = 0.1

.

The component of weight along the incline:

m g sin 30° = 5 \times 10 \times 0.5 = 25

N

The normal force:

N = m g cos 30° = 5 \times 10 \times \frac{3}{2} = 253

N

Friction force:

f_{k} = μ N = 0.1 \times 253 = 2.5 3

N

To move the block up:

F_{1} = m g sin 30° + f_{k} = 25 + 2.5 3

N

To prevent sliding down:

F_{2} = m g sin 30° - f_{k} = 25 - 2.5 3

∣ F_{1} ∣ - ∣ F_{2} ∣ = (25 + 2.5 3) - (25 - 2.5 3) = 53

N

The answer

53

N is not among the given options. The source marks this as BONUS, confirming none of the options is correct.

JEE Main · 2024 · 31 Jan 2024 · Shift 2 · Single correct option