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3 years ago

49.An object accelerates from rest to a speed of 24 m/s over a distance of 56 m. What

Physics

1 answer:

Nadusha1986 [10]3 years ago

3 0

<h3><u>Given</u><u>:</u></h3>

Initial velocity,u = 0 m/s

Final velocity,v = 24 m/s

Distance covered,s = 56 m

<h3><u>To</u><u> </u><u>be</u><u> </u><u>calculated:-</u><u> </u></h3>

Calculate the acceleration ( a ) .

<h3><u>Solution:-</u><u> </u></h3>

<u>According</u><u> </u><u>to</u><u> </u><u>the</u><u> </u><u>third</u><u> </u><u>equation</u><u> </u><u>of</u><u> </u><u>motion</u><u>, </u>

$\bf \: {v}^{2} = {u}^{2} + 2as$

★ Substituting the values in the third equation of motion:

$\sf \implies \: {(24)}^{2} = {(0)}^{2} + 2 \times a \times 56$

$\sf \implies \: 576 = 112a$

$\sf \implies \: a = \dfrac{576}{112}$

$\sf\implies \: a = 5.14 \: m {s}^{ - 2}$

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<h3>Data obtained from the question</h3>

From the question given above, the following data were obtained:

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2 years ago

In an experiment, a variable, position-dependent force F(x)F(x) is exerted on a block of mass 1.0kg1.0kg that is moving on a hor

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Answer:

The function F(x) for 0 < x < 5, the block's initial velocity, and the value of F(f).

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Given that,

Mass of block = 1.0 kg

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Frictional force = F(f)

Suppose, the following information would students need to test the hypothesis,

(A) The function F(x) for 0 < x < 5 and the value of F(f).

(B) The function a(t) for the time interval of travel and the value of F(f).

(D) The function a(t) for the time interval of travel, the time it takes the block to move 5 m, and the value of F(f).

(E) The block's initial velocity, the time it takes the block to move 5 m, and the value of F(f).

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The work done by a force is given by,

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Where, $F(x)$ = net force

We know, the net force is the sum of forces.

So, $\sum{F}=ma$

According to question,

We have two forces F(x) and F(f)

So, the sum of these forces are

$F(x)+(-F(f))=ma$

Here, frictional force is negative because F(f) acts against the F(x)

Now put the value in equation (I)

$W=\int_{x_{0}}^{x_{f}}{(F(x)-F(f))dx}$

We need to find the value of $\int_{x_{0}}^{x_{f}}{(F(x)-F(f))dx}$

Using newton's second law

$\int_{x_{0}}^{x_{f}}{(F(x)-F(f))dx}=\int_{x_{0}}^{x_{f}}{ma\ dx}$ ...(II)

We know that,

Acceleration is rate of change of velocity.

$a=\dfrac{dv}{dt}$

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$\int_{x_{0}}^{x_{f}}{(F(x)-F(f))dx}=\int_{v_{0}}^{v_{f}}{mv\ dv}$

$\int_{x_{0}}^{x_{f}}{(F(x)-F(f))dx}=\dfrac{mv_{f}^2}{2}+\dfrac{mv_{0}^2}{2}$

Now, the work done by the net force on the block is,

$W=\dfrac{mv_{f}^2}{2}+\dfrac{mv_{0}^2}{2}$

The work done by the net force on the block is equal to the change in kinetic energy of the block.

Hence, The function F(x) for 0 < x < 5, the block's initial velocity, and the value of F(f).

(C) is correct option.

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