Answer to Question #104308 in Classical Mechanics for Pichai

Question #104308
1. A car has a drag coefficient Cd = 0.30, a frontal area of A = 1.9 m^2 and a mass 1.2 tonnes. It has a coefficient of rolling resistance, Cr = 0.012. The rolling resistance is a force Fr = CrN, where N is the normal force. Hint: Retain accurate values until the final calculation, but remember significant figures.

i) What is Fr​ when traveling at v = 110 kph on a horizontal road?
ii) What power (in kilowatts) is required to overcome Fr​ at this speed?

2. A car has a drag coefficient Cd = 0.30, a frontal area of A = 1.9^2 m, a mass 1.2 tonnes and a coefficient of rolling resistance, Cr​, = 0.012. It is traveling up a hill with a slope of 1 in 20 at 110 kph. At what rate is it doing work against gravity (i.e. at what rate is it increasing its gravitational potential energy)?

Pg​= _____ kW.

A 1:20 grade means that it rises 1 m for every 20 m traveled along the road: sin⁡(θ)=1/20.
1
Expert's answer
2020-03-09T10:58:42-0400

1) i)

Fr=CrN=Crmg=(0.012)(1200)(9.8)=141 NF_r = C_rN=C_rmg=(0.012)(1200)(9.8)=141\ N

ii)


P=Frv=(141)1103.6=4300 W=4.3 kWP=F_rv=(141)\frac{110}{3.6}=4300\ W=4.3\ kW

2)


Pg=mgvcosα=mgvsinθP_g=mgv\cos{\alpha}=mgv\sin{\theta}

Pg=(1200)(9.8)1103.6120=18000 W=18 kWP_g=(1200)(9.8)\frac{110}{3.6}\frac{1}{20}=18000\ W=18\ kW


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