Question

Hi ! solve question given in attachment.​

Answer 1

Question :

A block of mass m slides along a floor while a force of magnitude F is applied to it at an angle $\tt \theta$ as shown in the figure.The coefficient of kinetic friction is $\tt {\mu}_{k}$.Then the blocks acceleration 'a' is given by : ( g is acceleration due to gravity )

Given :

mass : m kg

Force : F N

Coefficient of kinetic friction : $\tt {\mu}_{k}$

To Find :

acceleration of block (a) = ?

Diagram :

The free body diagram of the block is attached. Kindly find the attachment.

Solution :

Resolve force F into its two components

(i) component F$\tt cos\theta$ along horizontal.

(ii) component F$\tt sin\theta$ along vertical.

Let acceleration of block is directed towards right side.

Force of friction will oppose the motion of block and hence acts in opposite direction.

This force of friction is given by

f = $\tt {\mu}_{k}$N

Where, N:Normal Force

Now equating vertical forces we get,

N + Fsin$\theta$ = mg

$\tt \implies$ N = mg - Fsin$\theta$

$\therefore$ f = $\tt {\mu}_{k}$ ( mg - Fsin$\theta$ )

Now According to Newton's Second law

ma = Fcos$\tt \theta$ - f

$\tt \implies$ ma = Fcos$\tt \theta$ - $\tt {\mu}_{k}$ ( mg - Fsin$\theta$ )

$\tt \implies$ a = $\tt \frac{F}{m}cos\theta$ - $\tt {\mu}_{k}$ ( g - $\tt \frac{F}{m}sin\theta$ )

a = $\tt \frac{F}{m}cos\theta$ - $\tt {\mu}_{k}$ ( g - $\tt \frac{F}{m}sin\theta$ )

Option :

The correct option is option B)

Answer 2

Answer:

Given :

mass : m kg

Force : F N

Coefficient of kinetic friction : \tt {\mu}_{k}μ

k

To Find :

acceleration of block (a) = ?

Diagram :

The free body diagram of the block is attached. Kindly find the attachment.

Solution :

Resolve force F into its two components

(i) component F\tt cos\thetacosθ along horizontal.

(ii) component F\tt sin\thetasinθ along vertical.

Let acceleration of block is directed towards right side.

Force of friction will oppose the motion of block and hence acts in opposite direction.

This force of friction is given by

f = \tt {\mu}_{k}μ

k

N

Where, N:Normal Force

Now equating vertical forces we get,

N + Fsin\thetaθ = mg

\tt \implies⟹ N = mg - Fsin\thetaθ

\therefore∴ f = \tt {\mu}_{k}μ

k

( mg - Fsin\thetaθ )

Now According to Newton's Second law

ma = Fcos\tt \thetaθ - f

\tt \implies⟹ ma = Fcos\tt \thetaθ - \tt {\mu}_{k}μ

k

( mg - Fsin\thetaθ )

\tt \implies⟹ a = \tt \frac{F}{m}cos\theta

m

F

cosθ - \tt {\mu}_{k}μ

k

( g - \tt \frac{F}{m}sin\theta

m

F

sinθ )

a = \tt \frac{F}{m}cos\theta

m

F

cosθ - \tt {\mu}_{k}μ

k

( g - \tt \frac{F}{m}sin\theta

m

F

sinθ )

Option :

The correct option is option B)