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Maksim231197 [3]
3 years ago
9

Assuming that Albertine's mass is 60.0 kg , for what value of μk, the coefficient of kinetic friction between the chair and the

waxed floor, does she just reach the glass without knocking it over? Use g = 9.80 m/s2 for the magnitude of the acceleration due to gravity. g
Physics
1 answer:
makvit [3.9K]3 years ago
3 0

Answer:

\mu_k=0.101

Explanation:

It is given that,

Mass of Albertine, m = 60 kg

It can be assumed, the spring constant of the spring, k = 95 N/m

Compression in the spring, x = 5 m

A glass sits 19.8 m from her outstretched foot, h = 19.8 m

When she just reach the glass without knocking it over, a force of friction will also act on it. Using the conservation of energy for the spring mass system such that,

\dfrac{1}{2}kx^2=\mu_k mgh

\mu_k=\dfrac{kx^2}{2mgh}

\mu_k=\dfrac{95\times (5)^2}{2\times 60\times 9.8\times 19.8}

\mu_k=0.101

So, the coefficient of kinetic friction between the chair and the waxed floor is 0.101. Hence, this is the required solution.

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Find the magnitude of the resultant force and the angle it makes with the positive x-axis. (Let |a| = 22 lb and |b| = 16 lb. Rou
SVEN [57.7K]

Incomplete question as the angle between the force is not given I assumed angle of 55°.The complete question is here

Two forces, a vertical force of 22 lb and another of 16 lb, act on the same object. The angle between these forces is 55°. Find the magnitude and direction angle from the positive x-axis of the resultant force that acts on the object. (Round to one decimal places.)  

Answer:

Resultant Force=33.8 lb

Angle=67.2°

Explanation:

Given data

Fa=22 lb

Fb=16 lb

Θ=55⁰

To find

(i) Resultant Force F

(ii)Angle α

Solution

First we need to represent the forces in vector form

\sqrt{x} F_{1}=22j\\ F_{2}=u+v\\F_{2}=16sin(55)i+16cos(55)j\\F_{2}=16(0.82)i+16(0.5735)j\\F_{2}=13.12i+9.176j

Total Force

F=F_{1}+F_{2}\\ F_{2}=22j+13.12i+9.176j\\F_{2}=13.12i+31.176j

The Resultant Force is given as

|F|=\sqrt{x^{2} +y^{2} }\\|F|=\sqrt{(13.12)^{2} +(31.176)^{2} }\\ |F|=33.8lb

For(ii) angle

We can find the angle bu using tanα=y/x

So

tan\alpha =\frac{31.176}{13.12}\\ \alpha =tan^{-1} (\frac{31.176}{13.12})\\\alpha =67.2^{o}

7 0
3 years ago
a force is applied to a box of 10.0 kg for 4.0 s. the box goes from rest to 25 m/s in that time. What is the magnitude of that f
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Given:

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u(initial velocity)=0.(as the body is initially at rest).

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Now we know that

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Where v is the final velocity

u is the initial velocity

a is the acceleration acting on the body

t is the time of impact

Substituting these values we get

25=0+a x 4

4a=25

a=6.25m/s^2

Now we also know that

F=mxa

F=10 x6.25

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An automobile is sitting on a hill which is 20 m higher than ground level. Find the mass of the automobile if it contains 362,60
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