Question 29
The force resisting the motion of a car is proportional
to the square of the car’s speed. The magnitude of the force at a speed of 20.0
m s-1
is 800 N.
What useful output power is required from the car’s
engine to maintain a steady speed of
40.0 m s-1?
A 32
kW B
64 kW C
128 kW D
512 kW
Reference: Past Exam Paper – November 2018 Paper 11 Q17
Solution:
Answer:
C.
Resistive force F is
proportional to v2.
F ∝ v2
F = kv2 where
k is a constant
From the question, when force
F = 800 N when speed v = 20 m s-1.
k = F / v2 =
800 / 202 = 2
So, F = 2v2
The power output P = Fv
Since F = 2v2,
Output power P = Fv = (2v2)
× v = 2v3
When speed v = 40 m s-1,
Output power P = 2 × 403 = 128 000 W = 128 kW
thanks a lot, kindly assist me with Q5,Q11,Q13,Q23 and Q29 from the same paper 9702 pastexam paper-October/November paper 11 of 2018 (9702 w18 qp11)
ReplyDeletefor q23, go to
Deletehttp://physics-ref.blogspot.com/2019/06/the-diagram-shows-waveform-of-signal.html
thanks a lot, i have gotten the concept on this one.
DeleteYou are life saver man!!!
ReplyDelete2019 O/N P13 Q17
ReplyDeleteThe maximum useful output power of a car travelling on a horizontal road is P. The total resistive force acting on the car is kv^2, where v is the speed of the car and k is constant.
Which equation is correct when the car is travelling at maximum speed ?
A. v^3=P/k
B. v^2=P/k
C. v=(P/k)^2
D. v=(P/k)^3
go to
Deletehttp://physics-ref.blogspot.com/2020/06/the-maximum-useful-output-power-of-car.html