Question

A car of mass 100 kg moving on a horizontal road with a steady velocity of 10 m/s has a total frictional on it of 400N. Find the power due to the engine.

A car now climbs a hill an angle of 8° to the horizontal. Assuming frictional force stays same what engine power is needed to keep the car moving at Km s−1.

Answer 1

Answer on Question#49708 – Physics – Mechanics | Kinematics | Dynamics

1. $P_{1} = 4\frac{kJ}{s}$

2. $P_{2} = 5.364\frac{kJ}{s}$

Solution

1. Velocity $v = 10\frac{m}{s}$

Frictional force $F_{f} = 400\,N$

Time $t$

Distance $l = vt$

Power of engine $P$ is nothing else, but the energy required to compensate inhibitory effect of frictional force in unit time.

$$P_{1} = \frac{A_{1}}{t} = \frac{F_{f} l}{t} = F_{f} v$$ $$P_{1} = 400 \times 10 = 4000 \left(\frac{J}{s}\right) = 4 \left(\frac{kJ}{s}\right)$$

2. Angle $\alpha = 8{}^{\circ}$

Change of height $h$

Change of potential energy $\Delta U = mgh$

We have additional addendum in equation due to the change of car’s potential energy in Earth’s gravity field. Consider gravitational acceleration $g$ constant $\left(g = 9.8\frac{m}{s^2}\right)$.

$$P_{2} = \frac{A_{2}}{t} = \frac{F_{f} l + \Delta U}{t} = \frac{F_{f} l + mgh}{t} = F_{f} v + mgv \sin(\alpha) = P_{1} + mgv \sin(\alpha)$$ $$P_{2} = 4000 + 100 \times 9.8 \times 10 \times \sin(8{}^{\circ}) \approx 4000 + 1364 = 5364 \left(\frac{J}{s}\right) = 5.364 \left(\frac{kJ}{s}\right)$$

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