Question

Determine the power required for a 1150-kg car to climb a 100-m-long uphill road with a slope of 308 (from horizontal) in 12 s (a) at a constant velocity, (b) from rest to a final velocity of 30 m/s, and (c) from 35 m/s to a final velocity of 5 m/s. Disregard friction, air drag, and rolling resistance

Answer 1

Part a)

Power is defined as rate of work done

$P = \frac{W}{t}$

here we know that

$W = mgsin30 * d$

so power is given as

$P = \frac{mgsin30 * d}{t}$

$P = \frac{1150*9.8 * sin30 * 100}{12}$

$P = 46958.33 W$

Part b)

net work done on the block = change in kinetic energy

$W = \frac{1}{2}m(v_f^2 - v_i^2)$

$W = \frac{1}{2}*1150*(30^2 - 0^2) = 517500 J$

so power is given by

$P = \frac{517500}{12} = 43125 W$

Part c)

Part b)

net work done on the block = change in kinetic energy

$W = \frac{1}{2}m(v_f^2 - v_i^2)$

$W = \frac{1}{2}*1150*(5^2 - 35^2) = -690000 J$

so power is given by

$P = \frac{-690000}{12} = -57500 W$