Question

51. A car of mass 2000 kg is moving with a velocity
of 18 km/h. Work done to stop this car is
(A) 2.5 X 103 joule (B) 2.5 X 10 joule
(C) 2.5 X 105 joule (D) 2.5 x 104 joule​

Answer 1

Answer:

(D) 2.5 × 10^4 J

Explanation:

u = 18 km/hr = 18 × 5/18 = 5 m/s

work done = change in Kinetic energy

= |(1/2) m (v^2 - u^2)|

= (1/2) 2000 × 5 × 5

= 25000 J

Answer 2

Answer:

2.5 × 10⁴ joules

To find:

Work done to stop car.

Given:

mass of car (m) = 2,000 kg

velocity of car (u) = 18 km/hr

⇒ 18 × 5/18 = 5m/s

final velocity (v) = 0 m/s

Step-by-step explanation:

we know,

velocity is given by rate of change of displacement.

it is represented as,

v = Displacement/ time taken.

its unit is m/sec.

work done is given by object is,

⇒ change in kinetic velocity

⇒ $\frac{1}{2}mv^{2} - \frac{1}{2}mu^{2}$

Kinetic energy is the energy stored in the body due to its motion or velocity,

so, work done to stop car is

work done =  $\frac{1}{2}mv^{2} - \frac{1}{2}mu^{2}$

work done = $\frac{1}{2}m(v^{2} - u^{2})$

work done = $\frac{1}{2} [2000(5^{2} - 0^{2})$

work done = 1000 × 25

work done = 25,000

work done = 2.5 × 10⁴ joules

Hence,  2.5 × 10⁴ joules is work done to stop car.

Therefore, option (D) is correct.

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