Question

The speed of the vehicle of mass 500kg increase from 36 Km/hr to 72 Km/hr . Calculate the increase in its kinetic energy.

Answer 1

KE = k2-k1

=1/2m(v2-u2)

=250(400-100)

=75000 J

Answer 2

Answer:

If the speed of the vehicle of mass 500kg increases from 36 Km/hr to 72 Km/hr, then the increase in its kinetic energy is  $= 7.5 \times 10^4$J.

Explanation:

• Kinetic energy is the energy possessed by an object as a result of its motion. Essentially, it is the energy of mass in motion.
• Kinetic energy can never be negative, and it is a scalar quantity, meaning it only offers magnitude rather than direction.

K.E. = 1/2 mv²

where m mass of body and v is the velocity of the body.

Given v1= 36 Km/hr = 36 ×(5/18)  = 10 m/s

and v2 = 72 Km/hr  = 72×(5/18) = 20 m/s    

m = 500 kg

the increase in its kinetic energy=$k_2 - K_1$

$= \frac{1}{2} mv_1^2 - \frac{1}{2} mv_2^2$

$= \frac{1}{2} m[v_1^2 - v_2^2 ]$

$= \frac{1}{2} \times 500 [20^2 -10^2]\\ = \frac{1}{2} \times 500 [20-10](20+10)\\= \frac{1}{2} \times 500 (10)(30)$

= 250 [300 ]

= 75000J

$= 7.5\times 10^4$J

So, the increase in its kinetic energy is  $= 7.5 \times 10^4$J.