Question

A 10,000Kg car is traveling at 8m/s when it starts to climb a hill. Once the car gets to the top of the hill, it is traveling 5m/s. How much work was done by the car as it climbed the hill?​

Answer 1

Answer:

195000 J or 195 kJ

Explanation:

using work energy theorem ->  the net work done by the forces on an object equals the change in its kinetic energy

so ,

work done = change in kinetic energy

                  = initial - final kinetic energy

                  = $1/2 * 10000 * 8^{2} - 1/2 * 10000 * 5^{2}$

                  = 195000 J

Answer 2

Answer:

195000 J or 195 kJ

Explanation:

In work energy theorem, we know that,

=>

$work = kinetic \: energy$

$work = (initial \: KE - final \: KE)$

We know, KE = (mv²)/2

So,

$work = (10000 \times {8}^{2} \times \frac{1}{2} ) \: - \: (10000 \times {5}^{2} \times \frac{1}{2} )$

$work = 3,20,000 - 1,25,000$

$work = 1,95,000$

So, Work done by the car is 1,95,000J or 195kJ