JEE MAIN - Physics (2021 - 16th March Morning Shift - No. 3)
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)
_16th_March_Morning_Shift_en_3_1.png)
(g is acceleration due to gravity)
$${F \over m}\cos \theta + {\mu _K}\left( {g - {F \over m}\sin \theta } \right)$$
$${F \over m}\cos \theta - {\mu _K}\left( {g - {F \over m}\sin \theta } \right)$$
$$-$$$${F \over m}\cos \theta - {\mu _K}\left( {g - {F \over m}\sin \theta } \right)$$
$${F \over m}\cos \theta - {\mu _K}\left( {g + {F \over m}\sin \theta } \right)$$
Explanation
Drawing the FBD of the block.
_16th_March_Morning_Shift_en_3_2.png)
$$ \Rightarrow $$ N = mg $$-$$ Fsin$$\theta$$ ..... (1)
Also, Fcos$$\theta$$ $$-$$ $$\mu$$kN = m . a ..... (2)
Substituting the value of N from eq. (1) in eq. (2)
$$ \Rightarrow $$ Fcos$$\theta$$ $$-$$ $$\mu$$k(mg $$-$$ Fsin$$\theta$$) = m . a
$$ \Rightarrow $$ a = $${F \over m}$$cos$$\theta$$ $$-$$ $$\mu$$k$$\left( {g - {F \over m}\sin \theta } \right)$$
$$ \Rightarrow $$ N = mg $$-$$ Fsin$$\theta$$ ..... (1)
Also, Fcos$$\theta$$ $$-$$ $$\mu$$kN = m . a ..... (2)
Substituting the value of N from eq. (1) in eq. (2)
$$ \Rightarrow $$ Fcos$$\theta$$ $$-$$ $$\mu$$k(mg $$-$$ Fsin$$\theta$$) = m . a
$$ \Rightarrow $$ a = $${F \over m}$$cos$$\theta$$ $$-$$ $$\mu$$k$$\left( {g - {F \over m}\sin \theta } \right)$$
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