Question

please solve it by explanation​

Answer 1

Given :-

Weight of Truck = Force = mass = 1000 kg

Initial Velocity = 36 $kmh^{-1}$

New Velocity = 72 $kmh^{-1}$

To Find :-

Increase of Truck's Kinetic Energy

Explanation :-

• Initial Speed = 36 km/h = 36000 m / 3600s  = 10 m/s

• Final Speed = 72 km/h = 72000 m  / 3600s = 20 m/s

• Work Done = Increase of Kinetic Energy

⭐Using formula of $\frac{1}{2} mv^{2} - \frac{1}{2}mu^{2}$ We can find Increase of Kinetic Energy⭐

     Lets take $\frac{1}{2} m$ common, to find solution more easier and faster

⇒ $\frac{1}{2} m ( v^{2} -u^{2} )$

⇒ $\frac{1}{2}m (v+u) (v+u)$      -------------------- using formula $a^{2} -b^{2}$ = ( a + b ) ( a - b )

lets substitute value of m, v and u in expression

⇒ $\frac{1}{2}$  ×  1000 × ( 20 + 10 ) ( 20 - 10 )

⇒ $\frac{1}{2}$ × 1000 × 30 × 10

⇒ 500 × 300

⇒ 150000 Joules

∴ Increase of Kinetic Energy of Truck is 15 × $10^{4}$ J or  1.5 × $10^{5}$

Hope It Helps You ✌