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Natasha2012 [34]
3 years ago
5

Two crates, 3kg and 5 kg respectively, are connected with a thin rope.they are pulled up a rough plane which is inclined at 30de

gree to the horizontal.A force F is applied parallel to the plane as shown as in the diagram accelerating the system of crates at 2ms-2 up the slope.the frictional forces acting on 3kg and 5kg crates are 10N and 17N respectively. Draw a fully labeled free body diagram for each crate. calculate the component of the weight of each crate parallel l to the slope.calculate the magnitude of the applied force F and the tension T in the slope.

Physics
1 answer:
Makovka662 [10]3 years ago
6 0

Answer:

F = 85.24 N

T = 30.84 N

Explanation:

The parameters given are

Mass of crates = m₁ = 3 kg and m₂ = 5 kg

Component of masses acting along the plane = mgsin(θ)

Which gives;

F - (3×9.81×sin(30) + 10 + 5×9.81×sin(30) + 17) = m×a

So that we have;

2 = (F - 66.24)(3 + 5)

F = 16 + 66.24 = 85.24 N

The tension T between the crates = F × m₁/(m₁+m₂) = 85.24 × 3/(3+5) = 30.84 N

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Answer:

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3 years ago
A motor keep a Ferris wheel (with moment of inertia 6.97 × 107 kg · m2 ) rotating at 8.5 rev/hr. When the motor is turned off, t
Talja [164]

Answer:

P = 133.13 Watt

Explanation:

Initial angular speed of the ferris wheel is given as

\omega_i = 2\pi f

\omega_i = 2\pi(8.5/3600)

\omega_i = 0.015 rad/s

final angular speed after friction is given as

\omega_f = 2\pi f

\omega_f = 2\pi(7.5/3600)

\omega_f = 0.013 rad/s

now angular acceleration is given as

\alpha = \frac{\omega_f - \omega_i}{\Delta t}

\alpha = \frac{0.015 - 0.013}{15}

\alpha = 1.27 \times 10^{-4} rad/s^2

now torque due to friction on the wheel is given as

\tau = I \alpha

\tau = (6.97 \times 10^7)(1.27 \times 10^{-4})

\tau = 8875.3 N m

Now the power required to rotate it with initial given speed is

P = \tau \omega

P = 8875.3 \times 0.015

P = 133.13 Watt

8 0
3 years ago
3. What is the potential energy of a 8 Newton book sitting on a shelf that is 12 meters high?​
Svetradugi [14.3K]

Answer:

P = 96 J

Explanation:

Given that,

Weight of the book, W = mg = 8 N

It is placed at a height of 12 m

We need to find the potential energy of the book. The potential energy of an object is given by the formula as follows :

E = mgh

mg = Weight

E=8\ N\times 12\ m\\E=96\ J

So, the potential energy of the book is 96 J.

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4 years ago
How do metal detcors operate
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5 0
3 years ago
The spring is released and a 0.10-kilogram plastic sphere is fired from the launcher. Calculate the maximum speed with which the
tigry1 [53]

Answer:

A) The elastic potential energy stored in the spring when it is compressed 0.10 m is 0.25 J.

B) The maximum speed of the plastic sphere will be 2.2 m/s

Explanation:

Hi there!

I´ve found the complete problem on the web:

<em>A toy launcher that is used to launch small plastic spheres horizontally contains a spring with a spring constant of 50. newtons per meter. The spring is compressed a distance of 0.10 meter when the launcher is ready to launch a plastic sphere.</em>

<em>A) Determine the elastic potential energy stored in the spring when the launcher is ready to launch a plastic sphere.</em>

<em>B) The spring is released and a 0.10-kilogram plastic sphere is fired from the launcher. Calculate the maximum speed with which the plastic sphere will be launched. [Neglect friction.] [Show all work, including the equation and substitution with units.]</em>

<em />

A) The elastic potential energy (EPE) is calculated as follows:

EPE = 1/2 · k · x²

Where:

k = spring constant.

x = compressing distance

EPE = 1/2 · 50 N/m · (0.10 m)²

EPE = 0.25 J

The elastic potential energy stored in the spring when it is compressed 0.10 m is 0.25 J.

B) Since there is no friction, all the stored potential energy will be converted into kinetic energy when the spring is released. The equation of kinetic energy (KE) is the following:

KE = 1/2 · m · v²

Where:

m = mass of the sphere.

v = velocity

The kinetic energy of the sphere will be equal to the initial elastic potential energy:

KE = EPE = 1/2 · m · v²

0.25 J = 1/2 · 0.10 kg · v²

2 · 0.25 J / 0.10 kg = v²

v = 2.2 m/s

The maximum speed of the plastic sphere will be 2.2 m/s

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4 years ago
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