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Dmitry_Shevchenko [17]
2 years ago
15

A 6 kg block initially at rest is pulled to East along a horizontal, frictionless surface by a constant horizontal force of 12 N

. Find the speed of the block after it has moved 3m.
Physics
1 answer:
enot [183]2 years ago
4 0

Answers:

a) 3.46 m/s

b) 2 m/s^{2}

Explanation:

The complete question is written below:

A 6 kg block initially at rest is pulled to East along a horizontal, frictionless surface by a constant horizontal force of 12 N.

a) Find the speed of the block (using work-kinetic energy theorem) after it has moved 3.0 m.

b) Find the acceleration of the block using Newton.

<h3>a) Speed of the block after it has moved 3m</h3>

Let's begin by explaining the following:

The Work W done by a Force F refers to the release of potential energy from a body that is moved by the application of that force to overcome a resistance along a path.  In addition, W is a scalar quantity, and its unit in the International System of Units is the Joule (like energy).

Now, when the applied force is constant and the direction of the force and the direction of motion are parallel, the equation to calculate it is:  

W=F.d (1)

Where:

F= 12 N is the force exerted to pull the block to the East

d= 3 m is the distance the block has been moved

W=(12 N)(3 m) (2)

W=36 Nm= 36 J (3)

On the other hand, energy is the ability of matter to produce work in the form of movement, light, heat, among others. In this sense, according to the  <u>Work-Kinetic Energy Theorem</u> we have:

W=\Delta K=K_{f}-K_{o} (4)

Where:

\Delta K is the variation in kinetic energy K which is equal to:

K=\frac{1}{2}mV^{2} (5)

Where:

m=6kg is the mass of the block

V is the velocity

Hence, the <u>Work-Kinetic Energy Theorem </u> is written as follows:

W=\frac{1}{2}mV_{f}^{2} - \frac{1}{2}mV_{o}^{2} (6)

Where the initial velocity is zero (V_{o}=0 m/s) since we are told the block is initially at rest:

W=\frac{1}{2}mV_{f}^{2} (7)

So, we have to find the final velocity V_{f}, substituting (3) in (7):

36 J=\frac{1}{2}6 kg V_{f}^{2} (8)

V_{f}=3.46 m/s (9) This is the speed of the block after it has moved 3m

<h3>b) Acceleration of the block</h3>

According to <u>Newton's second law of motion, </u>which relates the force F with the mass m and the acceleration a of an object, we have:

F=m.a (10)

On the other hand, acceleration is the variation of velocity \Delta Vin time \Delta t:

a=\frac{\Delta V}{\Delta t} (11)

Then, equation (10) is rewritten as:

F=m \frac{\Delta V}{\Delta t} (12)

Isolating \Delta t:

\Delta t=\frac{m \Delta V}{F} (13)

Since \Delta V=V_{f}-V_{o} and V_{o}=0, we have:

\Delta V=V_{f}=3.46 m/s

\Delta t=\frac{(6 kg) (3.46 m/s)}{12 N} (14)

\Delta t=1.73 s (15)

Substituting (15) in (11):

a=\frac{3.46 m/s}{1.73 s} (16)

Finally:

a=1.99 m/s^{2} \approx 2 m/s^{2} This is the acceleration

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A car of mass 998 kilograms moving in the positive y–axis at a speed of 20 meters/second collides on ice with another car of mas
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Plz like if it helped.
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