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Sagot :
A. To drive the car on level ground, the engine needs to produce 11,500W of power.
b. The car's engine must provide 17,084W in power in order to move the vehicle up a hill with a 2.0 grade.
What is speed and how is it used?
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Briefing:
(a).The two forces at action are the drag force, D, which pushes the car backward, as well as the driving force, F, which pushes it vehicle forward. Since there can be no net force,
F-D=0
We know that D=500 N, so
F=D
F=500N
The force produced by the engine (F) and the vehicle's speed (v=23 m/s) are multiplied to produce the power used by the engine, so:
P=Fv
=(500N) (23m/s)
=11,500W
(b).the component of the weight of the car parallel to the slope, which acts against the motion of the car. Its magnitude is given by:
Wₓ=(mg)sin
=(710kg)(9.8m/s²)(sin2.0°)
=242.8N
F-D-Wx=0
F=D+Wx
=500N+242.8N
=248.8N
Consequently, this time the engine's power output
P=Fv
=(742.8N)(23m/s)
=17,084W
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The complete question is-
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 ∘ ?
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