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3 years ago

8

The coefficient of static friction between a book and the level surface it slides on is 0.65. If the mass of the book is 0.2 kg,

what minimum initial applied force is required to slide the books across the surface?

2 answers:

just olya [345]3 years ago

7 0

Answer: 1.274 N minimum force required to slide the book across the surface

Explanation:

Mass of the book = m = 0.2 kg

Normal force acting on the book,N = mg

Coefficient of friction = $mu_s$ = 0.65

Minimum force required to slide the book across the surface= F

$F=\mu_s\times N$

$F=0.65\times 0.2 kg\times 9.8 m/s^2=1.274 N$

1.274 N minimum force required to slide the book across the surface

klemol [59]3 years ago

3 0

Answer : F = 1.274 N

Explanation :

It is given that,

Coefficient of static friction, $\mu_s=0.65$

Mass of the book, $m=0.2\ kg$

We know that the relation between static friction and the force is :

$F=\mu_s\times N$

Where N is the normal force (N = mg)

So, $F=\mu_s\times m\times g$

$F=0.65\times 0.2\ kg\times 9.8\ m/s^2$

$F=1.274\ N$

Hence, the minimum initial applied force is required to slide the books across the surface is 1.274 N.

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2. In the activities below indicate if friction is useful or not useful? Explain your answer. i. Writing yes it is useful becaus

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Answer:

Please find the answer in the explanation

Explanation:

Friction is a force that opposes motion. One or two of the advantages of friction are break and ability of an object to walk.

Writing yes it is useful because when your writing because friction helps you see what your writing

ii. Rubbing. Yes, it is useful.

friction make it possible for two object to rub each other

iii. Skiing. No. It is not useful because With presence of friction, skiing will not be possible.

iv. Rotating a wheel No. It is not useful because Friction will oppose the rotation of the wheel.

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A spaceship negotiates a circular turn of radius 2925 km at a speed of 29960 km/h. (a) What is the magnitude of the angular spee

emmainna [20.7K]

a) 0.0028 rad/s

b) $23.68 m/s^2$

c) $0 m/s^2$

Explanation:

a)

When an object is in circular motion, the angular speed of the object is the rate of change of its angular position. In formula, it is given by

$\omega = \frac{\theta}{t}$

where

$\theta$ is the angular displacement

t is the time interval

The angular speed of an object in circular motion can also be written as

$\omega = \frac{v}{r}$ (1)

where

v is the linear speed of the object

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$v=29,960 km/h$ is the linear speed, converted into m/s,

$v=8322 m/s$

$r=2925 km = 2.925\cdot 10^6 m$ is the radius of the orbit

Subsituting into eq(1), we find the angular speed of the spaceship:

$\omega=\frac{8322}{2.925\cdot 10^6}=0.0028 rad/s$

b)

When an object is in circular motion, its direction is constantly changing, therefore the object is accelerating; in particular, there is a component of the acceleration acting towards the centre of the orbit: this is called centripetal acceleration, or radial acceleration.

The magnitude of the radial acceleration is given by

$a_r=\omega^2 r$

where

$\omega$ is the angular speed

$r$ is the radius of the orbit

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$\omega=0.0028 rad/s$ is the angular speed

$r=2925 km = 2.925\cdot 10^6 m$ is the radius of the orbit

Substittuing into the equation above, we find the radial acceleration:

$a_r=(0.0028)^2(2.925\cdot 10^6)=23.68 m/s^2$

c)

When an object is in circular motion, it can also have a component of the acceleration in the direction tangential to its motion: this component is called tangential acceleration.

The tangential acceleration is given by

$a_t=\frac{\Delta v}{\Delta t}$

where

$\Delta v$ is the change in the linear speed

$\Delta t$ is the time interval

In this problem, the spaceship is moving with constant linear speed equal to

$v=8322 m/s$

Therefore, its linear speed is not changing, so the change in linear speed is zero:

$\Delta v=0$

And therefore, the tangential acceleration is zero as well:

$a_t=\frac{0}{\Delta t}=0 m/s^2$

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