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m_a_m_a [10]
3 years ago
8

When being unloaded from a moving truck, a 16.0- kilogram suitcase is placed on a flat ramp inclined at 40.0 o. When released fr

om rest, the suitcase accelerates down the ramp at 1.36 m/s². What is the coefficient of kinetic friction between the suitcase and the ramp?
Physics
1 answer:
9966 [12]3 years ago
8 0

Answer:

0.65812

Explanation:

m = Mass of suitcase = 16 kg

\theta = Incline angle = 40°

g = Acceleration due to gravity = 9.81 m/s²

a = Acceleration of the block = 1.36 m/s²

f = Frictional force

The normal force is given by

N=mgcos\theta

In x direction

mgsin\theta-f=ma\\\Rightarrow f=mgsin\theta-ma\\\Rightarrow f=16\times 9.81\times sin40-16\times 1.36\\\Rightarrow f=79.13194\ N

Frictional force is given by

f=\mu N=\mu mgcos\theta\\\Rightarrow \mu=\dfrac{f}{mgcos\theta}\\\Rightarrow \mu=\dfrac{79.13194}{16\times 9.81\times cos40}\\\Rightarrow \mu=0.65812

The coefficient of kinetic friction between the suitcase and the ramp is 0.65812

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The component of the crate's weight that is parallel to the ramp is the only force that acts in the direction of the crate's displacement. This component has a magnitude of

<em>F</em> = <em>mg</em> sin(20.0°) = (15.0 kg) (9.81 m/s^2) sin(20.0°) ≈ 50.3 N

Then the work done by this force on the crate as it slides down the ramp is

<em>W</em> = <em>F d</em> = (50.3 N) (2.0 m) ≈ 101 J

The work-energy theorem says that the total work done on the crate is equal to the change in its kinetic energy. Since it starts at rest, its initial kinetic energy is 0, so

<em>W</em> = <em>K</em> = 1/2 <em>mv</em> ^2

Solve for <em>v</em> :

<em>v</em> = √(2<em>W</em>/<em>m</em>) = √(2 (101 J) / (2.0 m)) ≈ 10.0 m/s

3 0
3 years ago
A mass moves back and forth in simple harmonic motion with amplitude A and period T.
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a. 0.5 T

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During a full time period T, the mass on the spring oscillates back and forth, returning to its original position. This means that the total distance covered by the mass during a period T is 4 times the amplitude (4A), because the amplitude is just half the distance between the maximum and the minimum position, and during a time period the mass goes from the maximum to the minimum, and then back to the maximum.

So, the time t that the mass takes to move through a distance of 2 A can be found by using the proportion

1 T : 4 A = t : 2 A

and solving for t we find

t=\frac{(1T)(2 A)}{4A}=0.5 T

b. 1.25T

Now we want to know the time t that the mass takes to move through a total distance of 5 A. SInce we know that

- the mass takes a time of 1 T to cover a distance of 4A

we can set the following proportion:

1 T : 4 A = t : 5 A

And by solving for t, we find

t=\frac{(1T)(5 A)}{4A}=\frac{5}{4} T=1.25 T

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3 years ago
After landing on an unfamiliar planet, a space explorer constructs a simple pendulum of length 55.0 cmcm. She finds that the pen
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Hi there!

First, let's find the period of the pendulum. This can be found by solving for the amount of time it takes for the pendulum to make ONE complete swing.

T = \frac{\text{Total time}}{\text{Number of complete swings} }\\\\T = \frac{145}{110} = 1.318 s

Now, let's use the equation for the period of a simple pendulum:
T = 2\pi \sqrt{\frac{L}{g}}

T = Period (1.318 s)
L = length of string (0.55 m)

g = acceleration due to gravity on planet (? m/s²)

Let's solve for 'g' doing some quick rearranging of the equation:
T^2 = 4 \pi^2 (\frac{L}{g})\\\\g = \frac{4\pi^2 L}{T^2}\\\\

Solving for 'g' by plugging in values:
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3 0
2 years ago
A rectangular tank that is 4 feet long, 2 feet wide and 15 feet deep is filled with a heavy liquid that weighs 60 pounds per cub
V125BC [204]

Answer:

a) W₁ = 54000 Lb-ft

b) W₂ = 77760 Lb-ft

c)  W₃ = 24000 Lb-ft

W₄ = 40560 Lb-ft

Step by step

W= ∫₁² ydF         1  and  2 are the levels of liquid

Where dF is the differential of weight of a thin layer

y is the height of the differential layer and

ρ*V  = F

Then

dF = ρ* A*dy*g

ρ*g  =  60 lb/ft³

A= Area of the base then

Area of the base is:

A(b) = 4*2  =  8 ft²

Now we have the liquid weighs  60 lb/ft³

Then the work is:

a)

   W₁ = ∫₀¹⁵ 8*60*y*dy     ⇒   W₁ =480* ∫₀¹⁵ y*dy

W₁ =480* y² /2 |₀¹⁵      ⇒  480/2 [ (15)² - 0 ]

W₁ = 240*225

W₁ = 54000 Lb-ft

b) The same expression, but in this case we have to pump 3 feet higher, then:

W₂ = ∫₀¹⁸ 480*y*dy     ⇒ 480*∫₀¹⁸ydy    ⇒ 480* y²/2 |₀¹⁸

W₂ =  480/2 * (18)²

W₂ =  240*324

W₂ = 77760 Lb-ft

c) To pump two-thirds f the liquid we have

2/3* 15  =  10

W₃ = 480*∫₀¹⁰ y*dy   ⇒  W₃ = 480* y²/2 |₀¹⁰

W₃ = 240*(10)²

W₃ = 24000 Lb-ft

d)

W₄ =480*∫₀¹³ y*dy

W₄ =480* y²/2 |₀¹³

W₄ = 240*(13)²

W₄ = 240*169

W₄ = 40560 Lb-ft

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Answer:

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