A 710 kg car drives at a constant speed of 23 m/s . It is subject to a drag force of 500 N. What power is required from the car's engine to drive the car (a) on level ground? (b) up a hill with a slope of 2.0 ∘ ?
Added by Larry M.
Step 1
Substituting the given values, we get P = 500 N * 23 m/s = 11500 W or 11.5 kW. (b) When driving up a hill, the car not only needs to overcome the drag force, but also the gravitational force. The gravitational force can be calculated using the formula Fg = m * g Show more…
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$\mathrm{A} 710 \mathrm{kg}$ car drives at a constant speed of $23 \mathrm{m} / \mathrm{s}$. It is subject to a drag force of $500 \mathrm{N}$. What power is required from the car's engine to drive the car a. On level ground? b. Up a hill with a slope of $2.0^{\circ} ?$
Determine the power required for a 1150 -kg car to climb a 100 -m-long uphill road with a slope of $30^{\circ}$ (from horizontal) in $12 \mathrm{s}(a)$ at a constant velocity, $(b)$ from rest to a final velocity of $30 \mathrm{m} / \mathrm{s}$, and $(c)$ from $35 \mathrm{m} / \mathrm{s}$ to a final velocity of $5 \mathrm{m} / \mathrm{s}$. Disregard friction, air drag, and rolling resistance.
Determine the power required for a 1150-kg car to climb a 100-m-long uphill road with a slope of 30° (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.
Hubert A.
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