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 friction is μ k. Then, the block's acceleration 'a' is given by: (g is acceleration due to gravity) (a) F/mcos θ + μK(g - F/msin θ) (b) F/mcos θ - μK(g - F/msin θ) (c) -F/mcos θ - μK(g - F/msin θ) (d) F/mcos θ - μK(g + F/msin θ)

Answer: (b). Normal force: N = mg - Fsin θ (the force pulls upward at angle θ). Kinetic friction: f = μk(mg - Fsin θ) ma = Fcos θ - μk(mg - Fsin θ) a = F/mcos θ - μk(g - F/msin θ), so the answer is (b).

A block of mass m slides along a floor while a force of magnitude F… | JEE Main 2021 (16 Mar, Shift 1)

Q16JEE Main 2021 · 16 MAR · SHIFT 1Medium

Friction Basics

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)

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Answer: (b)

Normal force: $N = mg - F\sin\theta$ (the force pulls upward at angle $\theta$ ).
Kinetic friction: $f = \mu_k(mg - F\sin\theta)$
$ma = F\cos\theta - \mu_k(mg - F\sin\theta)$
$a = \dfrac{F}{m}\cos\theta - \mu_k\left(g - \dfrac{F}{m}\sin\theta\right)$ , so the answer is (b).

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