Question

an object of mass20 kg is moving with a velocity 15 m/s its velocity incresesto 35m/s on applaying this content force contiusly for 5s calculate theaccelaration and calculate the force applid​

Answer 1

Given :

Mass of object = 20kg

Initial velocity = 15m/s

Final velocity = 35m/s

Time interval = 5s

To Find :

Acceleration of object and applied force.

Solution :

❖ As per newton's second law of motion, force is defined as the rate of change of linear momentum.

• Force is a polar vector quantity as it has a point of application.
• SI unit : N

Formula : $\underline{\boxed{\bf{\orange{F=\dfrac{m(v-u)}{t}}}}}$

• F denotes force
• v denotes final velocity
• u denotes initial velocity
• t denotes time

By substituting the given values;

➙ F = 20 (35 - 15) / 5

➙ F = 20 (20) / 5

➙ F = 400/5

➙ F = 80 N

♦ Acceleration of the object :

➙ Acceleration = Force / Mass

➙ a = 80/20

➙ a = 4 m/s²

Answer 2

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$\star \: \bf\underbrace{Question} \: \star$

• An object of mass 20 kg is moving with a velocity 15 m/s, its velocity increases to 35m/s on applying this content force continuosly for 5s.Calculate the acceleration and calculate the force applied.

$\star \: \bf\underbrace{Given:-} \: \star$

• Mass of the object (m) = 20kg
• Initial velocity (u) = 15m/s
• Final velocity (v) = 35m/s
• Time taken (t) = 5s

$\star \: \bf\underbrace{To\:Find:-} \: \star$

• Acceleration of object and applied force.

To calculate:-

• Acceleration (a)
• Force applied (F)

$\text{\large\underline{\purple{Calculating Acceleration:- }}}$

As We know That,

• $\underline{\boxed{\bf{\red{a =\dfrac{v - u}{t}}}}}$

• $\sf { a = \dfrac{35-15}{5}}$

• $\sf { a = \dfrac{20}{5}}$

• ${\sf{\ a = 4m/{s}^{2}} }$

Therefore,

Acceleration of the object is -4m/s².

[Note : Negative sign implies that it is retardation.]

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$\text{\large\underline{\orange{Calculating Force applied:- }}}$

We know that,

• ${\large {\sf { F = ma}}}$

Where,

• F = force
• m = mass
• a = acceleration
• ${\large {\sf { F = 20 × 4}}}$
• ${\large {\sf { F = 80N}}}$

Therefore,

80N of force is applied on the object.

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