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 friction is \mu k. then, the block's acceleration 'a' is given by : (g is acceleration due to gravity)

F \over m\cos \theta - \mu _K( g - F \over m\sin \theta )

F \over m\cos \theta + \mu _K( g - F \over m\sin \theta )

- F \over m\cos \theta - \mu _K( g - F \over m\sin \theta )

F \over m\cos \theta - \mu _K( g + F \over m\sin \theta )

Laws of Motion PYQ — JEE Main 2021 · 16 Mar · Shift 1 · Q46 with Solution

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Correct answer: B

If the force $F$ is applied at angle $\theta$ above the horizontal, then:

Since the block does not accelerate vertically,

$N + F\sin\theta - mg = 0$

So,

$N = mg - F\sin\theta$

Kinetic friction opposes the motion, so its magnitude is

$f_k = \mu_k N = \mu_k (mg - F\sin\theta)$

Taking the direction of motion (and the horizontal pull) as positive:

$F\cos\theta - f_k = ma$

Substitute $f_k$ :

$F\cos\theta - \mu_k (mg - F\sin\theta) = ma$

$a = \frac{F\cos\theta - \mu_k (mg - F\sin\theta)}{m}$

$a = \frac{F}{m}\cos\theta - \mu_k\left(g - \frac{F}{m}\sin\theta\right)$

This matches Option B:

$\boxed{a = \frac{F}{m}\cos\theta - \mu_k\left(g - \frac{F}{m}\sin\theta\right)}$

Stored correct answer: B

My derived answer: B

So they agree.