Question

A 500kg car which was initially at rest traveled with an acceleration of 5m sec rise to power -2 its kinetic energy after 4 second was_ (using the formular for kinetic energy

Answer 1

Correct Question:

A 500 Kg car which was initially at rest traveled with an acceleration of 5 m/sec² .Its kinetic energy after 4 second was (using the formula for kinetic energy)

Answer:

• Kinetic Energy (K.E) is 100 Kilo Joules.

Given:

1. Mass of the Car (M) = 500 Kg.
2. Acceleration of the Car (a) = 5 m/sec².
3. Time Period (T) = 4 Second.

Explanation:

$\rule{300}{1.5}$

Applying First Kinematic Equation.

$\large \star \;{\boxed{\bold{v = u + at}}}$

$\bold{Here}\begin{cases}\text{v Denotes Final velocity} \\ \text{u Denotes Initial velocity} \\ \text{a Denotes Acceleration} \\ \text{t Denotes Time taken}\end{cases}$

$\large{\boxed{\tt v = u + at}}$

Substituting the Values.

$\large{\tt \hookrightarrow v = 0 + 5 \times 4}$

$\large{\tt \hookrightarrow v = 5 \times 4}$

$\large{\hookrightarrow {\underline{\underline{\tt v = 20 \; m/s}}}}$

$\rule{300}{1.5}$

$\rule{300}{1.5}$

From Kinetic energy,

$\large\star \; {\boxed{\bold{K.E = \dfrac{1}{2}Mv^2}}}$

$\bold{Here}\begin{cases}\text{M Denotes Mass} \\ \text{v Denotes Final Velocity}\end{cases}$

$\large{\boxed{\tt K.E = \dfrac{1}{2} Mv^2}}$

Substituting The values,

$\large{\tt \hookrightarrow K.E = \dfrac{1}{2} \times 500 \times (20)^2}$

$\large{\tt \hookrightarrow K.E = \dfrac{1}{2} \times 500 \times 400}$

$\large{\tt \hookrightarrow K.E = \dfrac{1}{\cancel{2}} \times 500 \times \cancel{400}}$

$\large{\tt \hookrightarrow K.E = 1 \times 500 \times 200}$

$\large{\tt \hookrightarrow K.E = 1 \times 100000}$

$\large{\tt \hookrightarrow K.E = 100000}$

$\huge{\boxed{\boxed{\tt K.E = 100 \; KJ}}}$

∴ Kinetic Energy of The car is 100 Kilo Joules.

$\rule{300}{1.5}$

Answer 2

$\huge{\underline{\underline{\red{\mathfrak{Answer :}}}}}$

$\large \tt Given \begin{cases} \sf{Mass \: of \: Car \: (m) \: = \: 5000 \: kg} \\ \sf{Acceleration \: (a) \: = \: 5 \: ms^{-2}} \\ \sf{Time \: interval \: (T) \: = \: 4 \: s} \end{cases}$

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To Find :

Kinetic Energy

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Solution :

Use Formula :

v = u + at

where,

• v is final velocity
• u is initial velocity
• a is acceleration t is time interval

Put Values ,

$\large \rightarrow {\sf{v \: = \: 0 \: + \: (5)(4)}}$

$\large \rightarrow {\sf{v \: = \: 20 \: ms^{-1}}}$

$\Large \leadsto {\underline{\boxed{\sf{v \: = \: 20 \: ms^{-1}}}}}$

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Now use formula for Kinetic Energy :

$\LARGE{\underline{\boxed{\sf{K.E \: = \: \frac{1}{2} \: mv^2}}}}$

Put Values

⇒K.E = ½ * 500 * (20)²

⇒K.E = ½*500*400

⇒K.E = ½*200000

⇒K.E = 100000

$\Large \leadsto {\underline{\boxed{\sf{K.E \: = \: 100 \: KJ}}}}$

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