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4vir4ik [10]
3 years ago
8

Based on Newton's law of motion, which combination of rocket bodies and engine will result in the acceleration of 40 m/s ^2 at t

he start of the launch?
Physics
1 answer:
ad-work [718]3 years ago
8 0

The question is incomplete. The complete question is :

The Rocket Club is planning to launch a pair of model rockets. To build the rocket, the club needs a rocket body paired with an engine. The table lists the mass of three possible rocket bodies and the force generated by three possible engines.

A 4-column table with 3 rows. The first column labeled Body has entries 1, 2, 3. The second column labeled Mass (grams) has entries 500, 1500, 750. The third column labeled Engine has entries 1, 2, 3. The fourth column labeled Force (Newtons) has entries 25, 20, 30.

Based on Newton’s laws of motion, which combination of rocket bodies and engines will result in the acceleration of 40 m/s2 at the start of the launch?

Body 3 + Engine 1

Body 2 + Engine 2

Body 1 + Engine 2

Body 1 + Engine 1

Solution :

Given :

Body       Mass (gram)     Engine      Force (newtons)

1                   500                 1                     25

2                  1500                2                    20

3                  750                  3                    30

The body 1 has a mass of 500 gram which is equal to 0.5 kg

And engine 2 has a force of 20 newtons.

We know that according to Newton's laws of motion,

Force = mass x acceleration

 20    = 0.5 x acceleration

Acceleration $=\frac{20}{0.5}$

                      $=\frac{200}{5}$

                      $= 40 \ m/s^2$

Therefore, based on laws of motion of Newton, the Body 1 + Engine 2 combination of the rocket bodies and engines will result in an acceleration of $ 40 \ m/s^2$ at the start of the launch.

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a. \displaystyle k_o=1350\ J

b. \displaystyle k_1=0\ J

c. \Delta k=-1350\ J

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<u>Work and Kinetic Energy </u>

When an object moves at a certain velocity v0 and changes it to v1, a change in its kinetic energy is achieved:

\Delta k=k_1-k_0

Knowing that

\displaystyle k=\frac{mv^2}{2}

We have

\displaystyle \Delta k=\frac{mv_1^2}{2}-\frac{mv_0^2}{2}

The work done by the force who caused the change of velocity (acceleration) is

\displaystyle W=\frac{mv_1^2}{2}-\frac{mv_0^2}{2}

If we know the distance x traveled by the object, the work can also be calculated by

W=F.x

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a. Before the slide, his initial kinetic energy is

\displaystyle k_o=\frac{mv_0^2}{2}

\displaystyle k_o=\frac{(75)6^2}{2}

\boxed{\displaystyle k_o=1350\ J}

b. Once he reaches the base, the player is at rest, thus his final kinetic energy is

\displaystyle k_1=\frac{(75)0^2}{2}

\boxed{\displaystyle k_1=0\ J}

c. The change of kinetic energy is

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\boxed{\Delta k=-1350\ J}

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\boxed{W=-1350\ J}

e. We compute the force of friction by using

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and solving for x

\displaystyle F=\frac{W}{x}

\displaystyle F=\frac{-1350\ J}{2\ m}

\boxed{F=-675\ N}

The negative sign indicates the force is against movement

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