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Question

An engine pulls a 1500kg car on a level road at a constant speed of 5.0m s1 against a frictional force of 500N. Calculate the power expended by the engine. What extra power has the engine to expend in order to maintain the same speed of the car up an inclined plane having a gradient of 1 in 10?

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Solution

Please refer work, power, energy related numericals
for this questions
wt.of car =1500Kg
constant speed of car =5.0ms1
then total force by the car to move for work =total force by engine + frictional force
=1500×5+500
f1=7500+500
f1=8000Nf1=8KN
Total work done =w1=F1×v=8000×5
W1=40,000watt
w1=40kw

case II
gradient 1 in 10 means
then sinθ=110.
Total Applied force =Mgsinθ+Frictionalforce
F2=1500×9.8×110+500 AB=10BC=1
F2=1970N

total work done w2=1970×5=9850walr
extra or 9.850Kw
Total work done w=w1+w2
=40+9.85=49.85

1350080_741580_ans_bcd8a1f71faf4554819c837babe8c9ed.PNG

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