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3 years ago

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A car accelerates uniformly from rest to 20 m/sec in 5.6 sec along a level stretch of road. Ignoring friction, determine the ave

rage power required to accelerate the car if (a) the weight of the car is 9,000 N, and (b) the weight of the car is 14,000 N.

1 answer:

bonufazy [111]3 years ago

3 0

Answer:

(a) $P=33000W$

(b) $P=51000W$

Explanation:

The average power is defined as the amount of work done during a time interval:

$P=\frac{W}{t}(1)$

According to work-energy theorem, the work done is equal to the change in kinetic energy. So, we have:

$W=\Delta K\\W=K_f-K_0\\W=\frac{mv_f^2}{2}-\frac{mv_0^2}{2}\\(2)$

Recall that the weight is given by:

$w=mg\\m=\frac{w}{g}(3)$

The car accelerates uniformly from rest ( $v_0=0$ ). Replacing (3) in (2), we have:

$W=\frac{wv_f^2}{2g}$

(a) Finally, we replace this in (1):

$P=\frac{wv_f^2}{2gt}\\P=\frac{9000N(20\frac{m}{s})^2}{2(9.8\frac{m}{s^2})(5.6s)}\\P=33000W$

(b)

$P=\frac{14000N(20\frac{m}{s})^2}{2(9.8\frac{m}{s^2})(5.6s)}\\P=51000W$

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