Question

1.a) What is the (average) power needed to accelerate a car of mass m=1550 kg from zero to 60 mph in 7.1 s?

b) Which simplifications did you make in that short calculation, i.e. what did you neglect?

2.The current BC hydro rates are 6.55 cents per kWh. If you have a desktop computer that uses 200 W (on average), how much does it cost you per month to run it for 6 hours per day?

3.

An elevator car has a mass of 1600 kg and is carrying passengers with a combined mass of 200 kg. A constant friction force of 4000 N is acting.

a) How much power must be delivered by the motor to lift the car with the passengers at a constant speed of 3m/s?

b) How much power must be delivered by the motor when the car with the passengers is moving with speed v = 3m/s, and it is supposed to provide an acceleration of 1.00m/s2?

Answer 1

1
given
m = 1550 kg
v = 60 mph = 60*1.609*5/18 = 26.82 m/s
time taken, t = 7.1 s
a)

Power needed to accelerate the car = kinetic energy gained by the car/time taken

= (1/2)*m*v^2/t

= (1/2)*1550*26.82^2/7.1

= 7.85*10^4 W <<<<<<--------------------Answer

b) I neglected air resistance and other dissipative forces.

2) Energy consumed in one month = power*time taken

= 200*6*60*60*30

= 1.296*10^8 J

= 1.296*10^8/(3*10^6) KWh (since 1 kWh = 3.6*10^6 J)

= 43.2 kWh

cost = 43.2*6.55

= 283 cents (or) $2.83 <<<<<<--------------------Answer

3)

Power delivred by the motor = F_motor*v

= (F_friction + weight of elevator and passengers)*v

= (4000 + 1600*9.8 + 200*9.8)*3

= 64920 W <<<<<<--------------------Answer

b)

Power delivred by the motor = F_motor*v

= (F_friction + (m_elevator + m_passengers)*(g + a) )*v

= (4000 + (1600 + 200)*(9.8 + 1) )*3

= 70320 W <<<<<<--------------------Answer