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Lera25 [3.4K]
3 years ago
6

In a pickup game of dorm shuffleboard,students crazed by final exams use a broom to propel a calculus book along the dorm hallwa

y. If the 3.5 kg book is pushed from rest through a distance of 0.90 m by the horizontal 25 N force from the broom and then has a speed of 1.60 m/s,what is the coefficient of kinetic friction between the book and floor?
Physics
1 answer:
Yakvenalex [24]3 years ago
6 0

Answer:

Coefficient of friction between the book and floor is 0.582.

Explanation:

Using the velocity formula;

v^2 = 2as

a = v^2/(2s)

a = 1.6^2/(2*0.9)

a = 2.56/1.8

a = 1.42 m/s^2

the force necessary to give the book the acceleration is  

F = ma = 3.5*1.42 (m is mass of the book i.e. 3.5 kg)

F = 4.98 N

The difference in the force is the friction force, which is

Ff = 25 - 4.98 = 20 N

Ff = mgμ

where μ is coefficient of friction and g is acceleration due to gravity that is 9.8 m/s^2

μ = Ff/mg

μ = 20/(3.5*9.81)

μ = 0.582

Coefficient of friction between the book and floor is 0.582.

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_____ friction is the force that sliding objects experience
Gnesinka [82]

Answer:

Sliding friction is the force that sliding objects experience

Explanation:

7 0
3 years ago
vector ????⃗ has a magnitude of 17.9 and its direction is 80∘ counter‑clockwise from the x- axis. what are the x- and y- compone
Reika [66]

We have vector (17.9*cos80^{0},17.9*sin80^{0})

Therefore,

x component = 17.9 * cos80 degree = 3.108

y component = 17.9 * sin80 degrees = 17.628

<h3>What is a vector?</h3>

An object with both magnitude and direction is referred to be a vector. A vector can be visualized geometrically as a directed line segment, with an arrow pointing in the direction and a length equal to the magnitude of the vector. The vector points in a direction from its tail to its head.

If the magnitude and direction of two vectors match, they are the same vector. This shows that if we move a vector to a different location without rotating it, the final vector will be the same as the initial vector. The vectors that denote force and velocity are two examples. The direction of force and velocity are both fixed. The size of the vector would represent the force's strength or the velocity's corresponding speed.

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5 0
2 years ago
Two workers are sliding 300 kg crate across the floor. One worker pushes forward on the crate with a force of 400 N while the ot
almond37 [142]

Answer:

The kinetic coefficient of friction of the crate is 0.235.

Explanation:

As a first step, we need to construct a free body diagram for the crate, which is included below as attachment. Let supposed that forces exerted on the crate by both workers are in the positive direction. According to the Newton's First Law, a body is unable to change its state of motion when it is at rest or moves uniformly (at constant velocity). In consequence, magnitud of friction force must be equal to the sum of the two external forces. The equations of equilibrium of the crate are:

\Sigma F_{x} = P+T-\mu_{k}\cdot N = 0 (Ec. 1)

\Sigma F_{y} = N - W = 0 (Ec. 2)

Where:

P - Pushing force, measured in newtons.

T - Tension, measured in newtons.

\mu_{k} - Coefficient of kinetic friction, dimensionless.

N - Normal force, measured in newtons.

W - Weight of the crate, measured in newtons.

The system of equations is now reduced by algebraic means:

P+T -\mu_{k}\cdot W = 0

And we finally clear the coefficient of kinetic friction and apply the definition of weight:

\mu_{k} =\frac{P+T}{m\cdot g}

If we know that P = 400\,N, T = 290\,N, m = 300\,kg and g = 9.807\,\frac{m}{s^{2}}, then:

\mu_{k} = \frac{400\,N+290\,N}{(300\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)}

\mu_{k} = 0.235

The kinetic coefficient of friction of the crate is 0.235.

5 0
3 years ago
A monkey has a bit of a heavy for on the gas pedal. As soon as the light turns green the monkey pushes the gas pedal to the floo
Andrei [34K]

Answer:

s=6.86m/s^2

Explanation:

Hello,

In this case, considering that the acceleration is computed as follows:

a=\frac{v_{final}-v_{initial}}{t}

Whereas the final velocity is 28.82 m/s, the initial one is 0 m/s and the time is 4.2 s. Thus, the acceleration turns out:

a=\frac{28.82m/s-0m/s}{4.2s}\\ \\s=6.86m/s^2

Regards.

3 0
3 years ago
The force required to stretch a Hooke’s-law
Setler [38]

I think this is correct, but I am not entirely certain.

Find the force constant of the spring:

F = - KX

(0 - 62.4) = -K(0.172m)

-362.791 = -K

362.791 N/m = K


Find the work done in stretching the spring:

W = (1/2)KX

W = (1/2)(362.791)(0.172m)

W = 31.2 J


5 0
3 years ago
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