What is the work done in KJ to increase the velocity of a car from 36 km - hr to 90 km - hr - When the weight of the car is 1000 kg-

v_2 = 90 * 5/18 = 25 m/sec, v_1 = 36 * 5/18 = 10m/sec Work done to increase the velocity will be = 1/2 m v_1^2 1/2 m v_2^2 = ( 25^2 10^2) 1000/2 = ...

What is the w...

Question

What is the work done (in KJ) to increase the velocity of a car from 36 $\frac{k m}{h r}$ to 90 $\frac{k m}{h r}$ ? When the weight of the car is 1000 kg.

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Solution

Given,
Final velocity, $v_{2}$ = 90 * $\frac{5}{18}$ = 25 $\frac{m}{s e c}$ ,

Initial velocity, $v_{1}$ = 36 * $\frac{5}{18}$ = 10 $\frac{m}{s e c}$

Work done to increase the velocity will be the change in kinetic energy.

$W = Δ K E = K E_{f} - K E_{i} = \frac{1}{2}$ m ${v_{2}}^{2}$ - $\frac{1}{2}$ m ${v_{1}}^{2}$

$W = (25^{2}$ - $10^{2}) \times \frac{1000}{2}$

$W = 500 \times (625 - 100) = 500 \times 525 = 262.5$ KJ.

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What is the work done (in KJ) to increase the velocity of a car from 36 kmhr to 90 kmhr? When the weight of the car is 1000 kg.

Given,Final velocity, v2 = 90 * 518 = 25 msec, Initial velocity, v1 = 36 * 518 = 10msec Work done to increase the velocity will be the change in kinetic energy. W=ΔKE=KEf−KEi=12 m v22 - 12 m v12 W=(252 - 102)×10002 W=500×(625−100)=500×525=262.5 KJ.

What is the work done (in KJ) to increase the velocity of a car from 36 $\frac{k m}{h r}$ to 90 $\frac{k m}{h r}$ ? When the weight of the car is 1000 kg.