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Colt1911 [192]
3 years ago
13

Imagine a 15kg block moving with a speed of 20m/s. Calculate the kinetic energy of this block. (Show the equation, show your wor

k and answer with units.)
Physics
2 answers:
sergeinik [125]3 years ago
8 0

Answer:

3000 Joules

Explanation:

Hello, I can help you with this.

the kinetic energy  of an object is the energy that it possesses due to its motion. It is defined as the work needed to accelerate a body of a given mass from rest to its stated velocity.it is given by:

E_{k}=\frac{1}{2}mv^{2}

Step 1

all you have to do is to put the values into the equation

Let

m= 15 kg

speed=20 m/s

E_{k}=\frac{1}{2}mv^{2}\\E_{k}=\frac{1}{2}(15kg)(20\frac{m}{s}) ^{2}\\E_{k}=\frac{1}{2}(15kg)(400)\frac{m^{2} }{s^{2} } \\E_{k}=3000 kg \frac{m^{2} }{s^{2} }

by definition

1Kg\frac{m^{2} }{s^{2} }=1 Joule

so, the kinetic energy of the block is 3000 Joules

I hope it helps, Have a great day.

dmitriy555 [2]3 years ago
5 0

Answer:

The answer to your question is:        Ke = 3000 Joules

Explanation:

Data

mass = 15 kg

speed = 20 m/s

Kinetic energy = ?

Equation

               Ke = \frac{1}{2}mv^{2}

               Ke = \frac{1}{2}(15)(20)^{2}

                Ke = 3000 Joules

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5) A 20.0 kg cart with no friction wheels sits on a table. A light string is attached to it and runs over a low friction pulley
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Answer:

1) Please find attached, created with Microsoft Visio

2) The acceleration of the masses connected by the light string is 0.00735 m/s²

3) The tension in the cord is 0.147 N

4) The time it would take the block to go 1.2 m to the edge of the table is approximately 18.07 s

5) The velocity of the cart as soon as it gets to the edge of the table is 0.042 m/s

Explanation:

1) Please find attached, the required free body diagram, showing the tension, weight and frictional (zero friction) forces acting on the cart and the mass created with Microsoft Visio

2) The acceleration of the masses connected by the light string is given as follows;

F = Mass, m × Acceleration, a

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The mass attached to the string, hanging rom the pulley, m = 0.0150 kg

The force, F acting on the system = The pulling force on the cart = The tension on the cable = The weight of the hanging mass = 0.0150 × 9.8 = 0.147 N

The pulling force acting on the cart, F = M × a

∴ F = 0.147 N = 20.0 kg × a

a = 0.147 N/(20.0 kg) = 0.00735 m/s²

The acceleration of the truck = a = 0.00735 m/s²

3) The tension in the cord = F = 0.147 N

4) The time, t, it would take the block to go 1.2 m to the edge of the table is given by the kinematic equation, s = u·t + 1/2·a·t²

Where;

s = The distance to the edge of the table = 1.2 m

u = The initial velocity = 0 m/s (The cart is assumed to be initially at rest)

a = The acceleration of the cart = 0.00735 m/s²

t = The time taken

Substituting the known values, gives;

s = u·t + 1/2·a·t²

1.2 = 0 × t + 1/2 ×0.00735 × t²

1.2 = 1/2 ×0.00735 × t²

t² = 1.2/(1/2 ×0.00735) ≈ 326.5306

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v² = u² + 2·a·s

u = 0 m/s

v² = 0² + 2 × 0.00735 × 1.2 = 0.001764

v = √(0.001764) = 0.042

The velocity of the cart as soon as it gets to the edge of the table = v = 0.042 m/s.

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