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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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My mass is 65 kg and on Earth this equals a weight of 640 N, but on the moon where gravity is 1.7 m/s² my
Helen [10]

Your weight on the moon given the data from the question is 110.5 N

<h3>Definition of mass and weight </h3>

Mass is simply defined as the quantity of matter present in an object. The mass of an object is constant irrespective of the location of the object.

Weight is simply defined as the gravitational pull on an object. The weight of an object varies from place to place due to gravity.

<h3>Relationship between mass and weight </h3>

Mass and weight are related according to the following equation

Weight (W) = mass (m) × Acceleration due to gravity (g)

<h3>How to determine the weight on the moon</h3>
  • Mass (m) = 65 Kg
  • Acceleration due to gravity on the moon (g) = 1.7 m/s²
  • Weight (W) =?

W = mg

W = 65 × 1.7

W = 110.5 N

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4 0
2 years ago
A rocket sled is tested at "5 g" (5 times the acceleration due to gravity). If the sled starts from rest at position do= 0.00, h
stealth61 [152]
D=at²
441m=(5*9.81m/s²)(t²)
t²=441/(5*9.81)
t≈√8.99
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8 0
3 years ago
Read 2 more answers
A 2000 kg car experiences a braking force of 10000N and skids to a 14 m stop. What was the speed of the car just before the brak
Fudgin [204]

Answer:

V = 11.83 m/s

Explanation:

Given the following data;

Mass = 2000 kg

Force = 10000N

Distance = 14 m

To find the final velocity of the car;

First of all, we would determine the acceleration of the car;

Acceleration = force/mass

Acceleration = 10000/2000

Acceleration = 5 m/s²

Next, we would use the third equation of motion to find the final velocity;

V^{2} = U^{2} + 2aS

Where;

V represents the final velocity measured in meter per seconds.

U represents the initial velocity measured in meter per seconds.

a represents acceleration measured in meters per seconds square.

S represents the displacement measured in meters.

Substituting into the equation, we have;

V² = 0² + 2*5*14

V² = 0 + 140

V = √140

V = 11.83 m/s

5 0
3 years ago
5. Dry ice is an example of _________, which is the process of a solid turning directly into a gas. (1 point) precipitation melt
ki77a [65]
<span>5. Dry ice is an example of _________, which is the process of a solid turning directly into a gas. (1 point)

sublimation


6. The ____ is a unit of force. (1 point)

</span>n<span>ewton

7. Which of the following is the boiling point of water? (1 point)
 
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8. Which of the following describes the molecular structure of water at 40°C? (1 point)

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6 0
3 years ago
Two friction disks A and B are brought into contact when the angular velocity of disk A is 240 rpm counterclockwise and disk B i
mel-nik [20]

Answer:

a) αA = 4.35 rad/s²

αB = 1.84 rad/s²

b) t = 3.7 rad/s²

Explanation:

Given:

wA₀ = 240 rpm = 8π rad/s

wA₁ = 8π -αA*t₁

The angle in B is:

\theta _{B} =4\pi =\frac{1}{2} \alpha _{B} t_{1}^{2}  =\frac{1}{2} (\frac{r_{A} }{r_{B} } )^{3} \alpha _{A} t_{1}^{2}=\frac{1}{2} (\frac{0.15}{0.2} )^{3} \alpha _{A} t_{1}^{2}

\alpha _{A} =8\pi (\frac{0.2}{0.15} )^{3} =59.57rad

w_{B,1} =\alpha _{B} t_{1}=(\frac{0.15}{0.2} )^{3} \alpha _{A} t_{1}=0.422\alpha _{A} t_{1}

The velocity at the contact point is equal to:

v=r_{A} w_{A} =0.15*(8\pi -\alpha _{A} t_{1})=1.2\pi -0.15\alpha _{A} t_{1}

v=r_{B} w_{B} =0.2*(0.422\alpha _{A} t_{1})=0.0844\alpha _{A} t_{1}

Matching both expressions:

1.2\pi -0.15\alpha _{A} t_{1}=0.0844\alpha _{A} t_{1}\\\alpha _{A} t_{1}=16.09rad/s

b) The time during which the disks slip is:

t_{1} =\frac{\alpha _{A} t_{1}^{2}}{\alpha _{A} t_{1}} =\frac{59.574}{16.09} =3.7s

a) The angular acceleration of each disk is

\alpha _{A}=\frac{\alpha _{A} t_{1}}{t_{1} } =\frac{16.09}{3.7} =4.35rad/s^{2} (clockwise)

\alpha _{B}=(\frac{0.15}{0.2} )^{3} *4.35=1.84rad/s^{2} (clockwise)

6 0
3 years ago
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