Question

answer please immediately​

Answer 1

1 m/s² is the required answer.

GIVEN:

• Mass of block, m = 5 kg

• Coefficient of friction of rough surface is 0.5

• Applied force , F = 30 N

TO FIND :

Acceleration of the block , a

SOLUTION:

Refer to the attachment

Normal force , N = mg

N = 5 × 10

N = 50 N

1. For the body to move, it must overcome the frictious force.
2. Frictious force must be lower than the applied force.
3. By calculation , the frictious force is estimated to be 25 N . [ f = 0.5 × 50 = 25 N]
4. So the body is able to move with some acceleration.
5. Frictious force acts in direction opposite to the applied force
6. Pseudo force acts in opposite direction to that of acceleration.
7. Equating forces on horizontal direction, we get the required acceleration.

ma + 25 = 30

5a = 30 - 25

5a = 5

a = 1 m/s²

ANSWER:

The acceleration of block is 1 m/s²

Answer 2

11th/Physics

Newton's law of motion

Answer :

Mass of block (M) = 5kg

Coefficient of friction (μ) = 0.5

Horizontal force (F) = 30N

Acc. due to gravity (g) = 10m/s²

We have to find net acceleration of the block$.$

Diagram :

$\setlength{\unitlength}{1mm}\begin{picture}(5,5)\put(0,0){\framebox(15,15){\sf{M}}}\put(-20,-0.1){\line(1,0){55}}\multiput(-19,0)(3,0){19}{\qbezier(0,0)(-1,-1)(-2,-2)}\put(15,7.5){\vector(1,0){15}}\put(31.3,6){\sf{F}}\put(7.5,0){\vector(0,-1){15}}\put(4,-20){\sf{Mg}}\put(7.5,15){\vector(0,1){15}}\put(6.3,31.5){\sf{N}}\put(0,0.2){\vector(-1,0){15}}\put(-17,1.3){\sf{f}}\end{picture}$

★ In this case, Frictional force opposes the motion of block because it acts in the opposite direction that of applied force.

First of all we need to find net force acts on the block.

⧪ Fnet = F - f

Max. static frictional force,

• f = μN

Where, N denotes normal reaction

⧪ Fnet = F - μN

In vertical equilibrium, N = mg

➠ Fnet = F - μmg

➠ Fnet = 30 - (0.5×5×10)

➠ Fnet = 30 - 25

➠ Fnet = 5N

Only net force will produce horizontal acceleration in the block. so we can say that,

➤ Fnet = ma

➤ 5 = 5a

➤ a = 5/5

➤ a = 1 m/s²

Cheers!