JEE PYQ: Laws Of Motion Question 1
Question 1 - 2021 | 16 Mar Shift 1] (MCQ)
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)
(1) $\frac{F}{m}\cos\theta - \mu_K\left(g - \frac{F}{m}\sin\theta\right)$
(2) $\frac{F}{m}\cos\theta - \mu_K\left(g - \frac{F}{m}\sin\theta\right)$ [Correct]
(3) $\frac{F}{m}\cos\theta - \mu_K\left(g + \frac{F}{m}\sin\theta\right)$
(4) $\frac{F}{m}\cos\theta + \mu_K\left(g - \frac{F}{m}\sin\theta\right)$
Show Answer
Answer: (2)
Solution
Normal force: $N = mg - F\sin\theta$. Along horizontal: $F\cos\theta - \mu_K N = ma$. So $F\cos\theta - \mu_K(mg - F\sin\theta) = ma$. Therefore $a = \frac{F}{m}\cos\theta - \mu_K\left(g - \frac{F}{m}\sin\theta\right)$.
BETA
