Question 116: Laws of Motion

Question

A block of mass m slides along a floor while a force of magnitude F is applied to it at an angle $$	heta$$ 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)

Accepted Answer

Drawing the FBD of the block.
$$ \Rightarrow $$ N = mg $$-$$ Fsin$$	heta$$ ..... (1)
Also, Fcos$$	heta$$ $$-$$ $$\mu$$
k
N = m . a ..... (2)
Substituting the value of N from eq. (1) in eq. (2)
$$ \Rightarrow $$ Fcos$$	heta$$ $$-$$ $$\mu$$
k
(mg $$-$$ Fsin$$	heta$$) = m . a
$$ \Rightarrow $$ a = $${F \over m}$$cos$$	heta$$ $$-$$ $$\mu$$
k
$$\left( {g - {F \over m}\sin 	heta } ight)$$

Answer

$${F \over m}\cos 	heta  + {\mu _K}\left( {g - {F \over m}\sin 	heta } ight)$$

Answer

$${F \over m}\cos 	heta  - {\mu _K}\left( {g - {F \over m}\sin 	heta } ight)$$

Answer

$$-$$$${F \over m}\cos 	heta  - {\mu _K}\left( {g - {F \over m}\sin 	heta } ight)$$

Answer

$${F \over m}\cos 	heta  - {\mu _K}\left( {g + {F \over m}\sin 	heta } ight)$$

Laws of Motion

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)

Choose the correct answer: