Question

The work required to be done to stop a car of mass 1500 kg moving at a velocity of 90 km/h

Answer 1

$\huge\textbf{Answer}$


Mass (m) = 1500 Kg

Final Velocity (v) = 90 km/h

= 90 × $\dfrac{5}{18}$

= 25 m/s

Initial Velocity (u) = 0

We have to find work done required to stop the car.

Means change in kinetic Change = $\dfrac{1}{2}$ mv² - $\dfrac{1}{2}$ mu²


= ½ m (v² - u²)

= ½ × 1500 [(25)² - (0)²]

= 750 × 625

= 468750 J

Answer 2

$\huge\mathfrak{Answer}$

m=1500kg

u= 60km/h

=60* 5/18

= 50/3m/s

v =0m/s w = change in its KE.

initial KE =1/2 mu2 final

KE=1/2mv² w = 1/2mv2 - 1/2 mu2

=1/2m(v2-u2)

= 1/2 * 15000{0- (50/3)2}

= -208333.33J

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