Question

A constant power P is applied to a car of mass M = 3000 kg. The velocity of the car increases from v = 2 ms-1 to v = 5 ms -1 when car travels a distance of 117 m. The value of power P is​

Answer 1

Answer:

942.3 W

Explanation:

Mass, m = 3000 kg

Initial Velocity, u = 2 m/s

Final Velocity, v = 5 m/s

Distance, s = 117 m

(v^2 - u^2)/2s = a

Acceleration, a = (25 - 4)÷(2 × 117) = 21/234 = 7/78 m/s^2

Force = ma = 3000×7/78 N = 21000/78 N

Work = F × s = 21000/78 × 117 J = 2,457,000/78 J

Time, t = (v - u)÷a = 3×78/7 = 234/7 s

Power = Work ÷ Time = 2,457,000/78 × 7/234 W = 942.3 W

Answer 2

Answer:

The value of power is 1000 Watt .

Explanation:

Constant power, $P = F V = maV$                      

So,we get

   $a= \frac{P}{mV}$

  → $V \frac{dV}{dS} = \frac{P}{mV}$

  →  $V^2 \frac{dV}{dS} = \frac{P}{m}$

  → $\frac{P}{m} dS = V^2 \frac{dV}{dS}$

Taking integration on both side,we will get

      $\int\limits^{117}_0{\frac{P}{m} dS = \int\limits^5_2 V^2 dV$

→ $\frac{P}{m} \times 117 = \frac{1}{3} [5^3 - 2^3]$

→ $\frac{P}{m} \times 117 = \frac{1}{3} \times 117$

→ $P = \frac{m}{3}$ $= \frac{3000}{3} = 1000$

∴ The value of power is 1000 Watt .