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erica [24]
3 years ago
15

A 60 kg block slides along the top of a 100 kg block with an acceleration of 2.0 m/s2 when a horizontal force F of 340 N is appl

ied. The 100 kg block sits on a horizontal frictionless surface, but there is friction between the two blocks.
Physics
1 answer:
Taya2010 [7]3 years ago
3 0

Answer:

The coefficient of friction and acceleration are 0.37 and 2.2 m/s²

Explanation:

Suppose we find the coefficient of friction and the acceleration of the 100 kg block during the time that the 60 kg block remains in contact.

Given that,

Mass of block = 60 kg

Acceleration = 2.0 m/s²

Mass = 100 kg

Horizontal force = 340 N

Let the frictional force be f.

We need to calculate the frictional force

Using balance equation

F-f=ma

Put the value into the formula

340-f=60\times2.0

f=340-60\times2.0

f=220\ N

We need to calculate the coefficient of friction

Using formula of friction force

f= \mu mg

\mu=\dfrac{f}{mg}

\mu=\dfrac{220}{60\times9.8}

\mu =0.37

We need to calculate the acceleration of the 100 kg block

Using formula of newton's law

F = ma

a=\dfrac{F}{m}

a=\dfrac{220}{100}

a=2.2\ m/s^2

Hence, The coefficient of friction and acceleration are 0.37 and 2.2 m/s²

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The force needed to give a car of mass 800 kg an acceleration of 2.0 ms-² is 1600N.

<h3>How to calculate force?</h3>

The force needed to push an object can be calculated by multiplying the mass of the object by its acceleration as follows:

Force = mass × acceleration

According to this question, a car of mass 800 kg has an acceleration of 2.0 ms−². The force is calculated as follows:

Force = 800kg × 2m/s²

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Therefore, the force needed to give a car of mass 800 kg an acceleration of 2.0 ms-² is 1600N.

Learn more about force at: brainly.com/question/13191643

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Explanation:

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When you do what you have described, you are setting a stage that not even the USS Enterprise (Star Trek) can get out of. The increase is huge.

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Because this is in the denominator, the 1/4 is going to flip to the numerator.

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Think about what that means. If you were out golfing, your drives would be roughly 1/16 times as far as they are now. Also you would be lugging around 16 times your weight around the golf course. My feeling is that you would never finish 5 holes at that rate.

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