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ludmilkaskok [199]
3 years ago
14

A horizontal force of 12N is applied to 1.5kg of block which rests on a horizontal surface. If the coefficient friction is 0.3,

find the acceleration produced in the block. (take g=10m/s^2).
Class 11 Physics
Please answer asap!! ​
Physics
1 answer:
Rama09 [41]3 years ago
8 0

Answer:

7 m/s²

Explanation:

From the question given above, the following data were:

Force applied (Fₐ) = 15 N

Mass (m) of block = 1.5 Kg

Acceleration due to gravity (g) = 10 m/s²

Coefficient of friction (μ) = 0.3

Acceleration (a) of block =?

Next, we shall determine the frictional force. This can be obtained as follow:

Mass (m) of block = 1.5 Kg

Acceleration due to gravity (g) = 10 m/s²

Coefficient of friction (μ) = 0.3

Normal reaction (R) = mg = 1.5 × 10 = 15 N

Frictional force (Fբ) =?

Fբ = μR

Fբ = 0.3 × 15

Fբ = 4.5 N

Next, we shall determine the net force acting on the block. This can be obtained as follow:

Force applied (Fₐ) = 15 N

Frictional force (Fբ) = 4.5 N

Net force (Fₙ) =.?

Fₙ = Fₐ – Fբ

Fₙ = 15 – 4.5

Fₙ = 10.5 N

Finally, we shall determine the acceleration produced in the block. This can be obtained as follow:

Mass (m) of block = 1.5 Kg

Net force (Fₙ) = 10.5 N

Acceleration (a) of block =?

Fₙ = ma

10.5 = 1.5 × a

Divide both side by a

a = 10.5 / 1.5

a = 7 m/s²

Therefore, the acceleration produced in the block is 7 m/s²

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The component of the force in negative z-direction is -0.144 N.

The given parameters;

  • <em>current in the wire, I = 2.7 A</em>
  • <em>length of the wire, L = (3.2 i + 4.3j) cm</em>
  • <em>magnetic filed, B = 1.24 i</em>

The force on the segment of the wire is calculated as follows;

F = ILBsin(\theta)

where;

  • <em>θ is the angle wire and magnetic field</em>

<em />

The force on the wire segment will be perpendicular in negative z-direction (applying right hand rule), so there won't be any x and y component of the force.

The angle between the wire and the magnetic field is calculated as follows;

\theta = tan^{-1} (\frac{y}{x} )\\\\\theta = tan^{-1} (\frac{4.3}{3.2} )\\\\\theta = 53.3 \ ^0

The magnitude of the wire length is calculated as follows;

|l | = \sqrt{3.2^2 + 4.3^2} = 5.36 \ cm = 0.0536 \ m

The component of the force in negative z-direction is calculated as;

F_z = -ILB sin(\theta)\\\\F_z = -2.7 \times 0.0536 \times 1.24 \times  sin(53.3)\\\\F_z = -0.144 \ N

Thus, the component of the force in negative z-direction is -0.144 N.

Learn more here:brainly.com/question/22719779

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