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What is the KE of a 4500kg at a speed of 20 m/s?

Sagot :

[tex] \\ \maltese \: \: \: \: \large{\underline{\underline{\pmb{ \textbf{\textsf{\color{purple}{Given \: :}}}}}} } \\ \\ [/tex]

  • Mass ( m ) of the body = 4,500 kg
  • velocity of the body = 20 m/s

[tex] \\ \maltese \: \: \: \: \large{\underline{\underline{\pmb{ \textbf{\textsf{\color{green}{To \: Find \: :}}}}}} } \\ \\ [/tex]

  • kinetic energy (ke) = ??

[tex] \\ \maltese \: \: \: \: \large{\underline{\underline{\pmb{ \textbf{\textsf{\color{aqua}{Solution \: :}}}}}} } \\ \\ [/tex]

Calculating the kinetic energy;

[tex] \\ \maltese \: \: \: \: \large{\underline{\underline{\pmb{ \textbf{\textsf{\color{hotpink}{using \: formula \: :}}}}}} } \\ \\ [/tex]

[tex] \red\bigstar { \underline{ \underline{\underline{\boxed{\color{blue}{\sf{ \:Kinetic \: \: energy \: = \frac{1}{2} \: \: m \: v \: {}^{2} }}}}}} }\\ \\ [/tex]

Where ,

  • (Ke) = kinetic energy.
  • m = mass
  • v = velocity

[tex] \\\\\sf \implies \: \:Kinetic \: \: energy \: = \frac{1}{2} \: \: m \: v \: {}^{2} \\ [/tex]

[tex] \\\sf \implies \: \:Kinetic \: \: energy \: = \frac{1}{2} \: \times \: 4,500 \: \times \: { 20}^{2} \\ [/tex]

[tex] \\\sf \implies \: \:Kinetic \: \: energy \: = \frac{1}{2} \: \times \: 4,500 \: \times \: 400\\[/tex]

[tex] \\\sf \implies \: \:Kinetic \: \: energy \: = \frac{1}{ \cancel{2}} \: \times \: \cancel {4,500 }\: \times \: 400\\[/tex]

[tex]\\ \sf \implies \: \:Kinetic \: \: energy \: = 2,250 \: \times \: 400\\[/tex]

[tex] \\\bf \implies{ \pink{ \: \:Kinetic \: \: energy \: = 9,00,000 \: \: \: joules}}\\\\\\[/tex]

hence the kinetic energy (ke) is 9,00,000 Joules...!!

Hi1315

Answer:

900kJ

Explanation:

The formula to find the kinetic energy is :

[tex]E_k=\frac{1}{2} mv^{2}[/tex]

Here,

m ⇒ mass ⇒ 4500 kg

v ⇒ velocity ⇒ 20 m/s

Let us find it now.

[tex]E_k=\frac{1}{2} mv^{2}[/tex]

[tex]E_k=\frac{1}{2} *4500*20^{2}[/tex]

[tex]E_k=\frac{1}{2} *1800000[/tex]

[tex]E_k=900000J[/tex]

[tex]E_k=900kJ[/tex]