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melisa1 [442]
3 years ago
11

A block of mass 2 kg is placed on the floor. The coefficient of static friction is 0.4. A horizontal force of 2.5 N is applied o

n the block. Calculate the force of friction between the block and the floor.​
Physics
1 answer:
almond37 [142]3 years ago
5 0

Answer: 2.5N

Explanation:

Given the following :

Mass of block (m) = 2kg

Coefficient of static friction (μs) = 0.4

Horizontal force applied to the block = 2.5N

The frictional force (Ff) between the block and the floor is :

First calculate the maximum static frictional force:

Frictional force = Coefficient of static friction(μs) × normal reaction(R)

Normal reaction(R) = mass × acceleration due to gravity (10m/s^2)

R = 2 × 10 = 20

Fmax = μs × R

Fmax = 0.4 x 20 = 8N

Here, since the applied force (2.5N) is less than maximum frictional force(8N).

The force of friction between the block and the floor will be equal to the applied force of 2.5N due its ability to adjust itself in other to ensure equilibrium.

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Answer:

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Explanation:

From the question we are told that

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Generally the frequency of the circular motion is  

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except ii and iii

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Earning Goal: To be able to calculate work done by a constant force directed at different angles relative to displacement
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Answer:

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8 0
2 years ago
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Answer:

Explanation:

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It implies that a force ( frictional ) acts on it which is equal to 76 N in opposite direction ( friction )

When we apply  a greater force on it it starts moving with acceleration .

This time kinetic friction acts on it due to rough ground equal to 76 N .This is limiting friction ( maximum friction )

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=

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igomit [66]

Answer;

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