Question

Hi ....in this question i am doing it with 1/2 m(v2-u2) simply...but Answer is not coming with this method.. please try to solve it with this method and tell me please​

Answer 1

Given:

Initial velocity (u) = 30 km/h = $\sf \dfrac{25}{3}$ m/s

Final velocity (v) = 60 km/h = $\sf \dfrac{50}{3}$ m/s

Mass of car (m) = 1500 kg

To Find:

Work done by the force on the car (W)

Answer:

WORK-ENERGY THEOREM

The change in kinetic energy of a particle is equal to the work done on it by the net force.

$\bf \leadsto W = \Delta K \\ \\ \rm \leadsto W = K_f - K_i \\ \\ \rm \leadsto W = \dfrac{1}{2} mv^2 - \dfrac{1}{2} mu^2 \\ \\ \bf \leadsto W = \dfrac{1}{2} m(v^2 - u^2) \\ \\ \rm \leadsto W = \dfrac{1}{2} \times 1 500\Bigg[ \bigg( \dfrac{50}{3} \bigg) ^2 - \bigg(\dfrac{25}{3} \bigg) ^2\Bigg] \\ \\ \rm \leadsto W = 750 \bigg( \dfrac{2500}{9} - \dfrac{625}{9} \bigg) \\ \\ \rm \leadsto W = 750 \times \dfrac{(2500 - 625)}{9} \\ \\ \rm \leadsto W = 750 \times \dfrac{1875}{9} \\ \\ \rm \leadsto W = 750 \times \dfrac{625}{3} \\ \\ \rm \leadsto W = 250 \times 625 \\ \\ \rm \leadsto W = 156250 \: J \\ \\ \rm \leadsto W = 156.25 \: kJ$

$\therefore$ Work done by the force on the car (W) = 156.25 kJ

Answer 2

Answer:

Mass of car=1500 kg

initial velocity of car, u=30km/h

=(30*1000)/(60*60)

=25/3 m/s

Similarly, the final velocity of the car,v=60km/h

=50/3

Thus, the work done=Change in kinetic energy

=1/2m(v²-u²)

=1/2*1500((50/3)²-(25/3)²)

=750((2500/9)-(625/9))

taking the LCM of the denominator

=750(2500-625)/9

=(750*1875)/9

=1406250/9

=156250