Question 139: 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)
![](https://imagex.cdn.examgoal.net/1kmili649/c5367306-4ff7-43e9-84e6-062048ffac1d/cb587690-89f4-11eb-9c26-d9a423ffb060/file-1kmili64a.jpg)

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
$\left( {g - {F \over m}\sin 	heta } ight)$
![](https://imagex.cdn.examgoal.net/1kp3tcrqp/f35adde9-6818-47c1-a54f-bfbf7b0b2657/629fa520-bd38-11eb-9b93-ddbbaf4f20e8/file-1kp3tcrqq.png)

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: