A block of mass $m$ slides along a floor while a force of magnitude F is applied to it at an angle $\theta$ as shown in figure. The coefficient of kinetic is $\mu_{\mathrm{K}}$. Then, the block's acceleration '$a$' is given by : ( g is acceleration due to gravity)

(C) $\frac{F}{m} \cos \theta-\mu_{\mathrm{K}}\left(g-\frac{F}{m} \sin \theta\right)$

(A) $\frac{\mathrm{F}}{\mathrm{m}} \cos \theta+\mu_{\mathrm{K}}\left(\mathrm{g}-\frac{\mathrm{F}}{\mathrm{m}} \sin \theta\right)$

(B) $\frac{F}{m} \cos \theta-\mu_{k}\left(g+\frac{F}{m} \sin \theta\right)$

(D) $-\frac{\mathrm{F}}{\mathrm{m}} \cos \theta-\mu_{\mathrm{K}}\left(\mathrm{g}-\frac{\mathrm{F}}{\mathrm{m}} \sin \theta\right)$

JEE Main · 2021 · 16 Mar 2021 · Shift 1 · Laws of Motion

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A block of mass

m

slides along a floor while a force of magnitude F is applied to it at an angle

θ

as shown in figure. The coefficient of kinetic is

μ_{K}

. Then, the block's acceleration '

a

' is given by :
( g is acceleration due to gravity)

(A) $\frac{F}{m} cos θ + μ_{K} (g - \frac{F}{m} sin θ)$
(B) $\frac{F}{m} cos θ - μ_{k} (g + \frac{F}{m} sin θ)$
(C) $\frac{F}{m} cos θ - μ_{K} (g - \frac{F}{m} sin θ)$ ✓ Correct
(D) $- \frac{F}{m} cos θ - μ_{K} (g - \frac{F}{m} sin θ)$

Answer: (C)

Solution

The applied force points above the horizontal, so its vertical component lifts the block and reduces how hard the floor presses back. Balancing the vertical forces,

N = m g - F sin θ

The block is sliding, so friction acts backwards with magnitude

f = μ_{K} N = μ_{K} (m g - F sin θ)

. Along the floor,

ma = F cos θ - μ_{K} (m g - F sin θ)

Dividing by

m

gives the acceleration

a = \frac{F}{m} cos θ - μ_{K} (g - \frac{F}{m} sin θ)

JEE Main · 2021 · 16 Mar 2021 · Shift 1 · Single correct option

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JEE Main · 2021 · 16 Mar 2021 · Shift 1 · Laws of Motion

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