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Integrated Concepts

(a) Assuming 95.0%efficiency for the conversion of electrical power by the motor, what current must the 12.0V batteries of a 750kg electric car be able to supply: (a) To accelerate from rest to 25.0m/ms s in 1.00min? (b) To climb a 2.00ร—102m high hill in 2.00min at a constant 25.0m/ms s speed while exerting 5.00ร—102N of force to overcome air resistance and friction? (c) To travel at a constant 25.0m/ms s speed, exerting a 5.00ร—102Nforce to overcome air resistance and friction? See Figure 20.44.

Short Answer

Expert verified
  1. The amount of current is 343A.
  2. The amount of current is 2.17kA.
  3. The amount of current is 1.10kA.

Step by step solution

01

Definition of Power:

The power of a process is the amount of some type of energy converted into a different type divided by the time intervalฮ”tin which the process occurred:

P=ฮ”Eฮ”t โ€ฆ.. (1)

The SI unit of power is the watt.

As you know, one watt is equal to one joule per second. Therefore,

1W=1J/Js s

The power of any electric circuit is defined by the formula

P=Iฮ”V โ€ฆ.. (2)

Here,Iis the current andIฮ”Vis the potential difference.

02

Definition of Kinetic Energy

The kinetic energy of any particle,

K.E=12mv2 โ€ฆ.. (3)

Wheremis the mass andvis the velocity of the particle.

03

Definition of Gravitational potential energy

The gravitational potential energy is given by

Ug=mgh โ€ฆ.. (4)

Here,mis the mass of the particle,gis the acceleration due to gravity andhis the height of the particle.

04

Definition of Work done:

The work done is given by

W=Fdcosฮธ โ€ฆ.. (5)

Here,Fis the exerted force,dis the displacement,ฮธis the angle betweenFโ†’anddโ†’.

05

Definition of average speed:

The average speed is

v=ฮ”xt......(6)

Whereฮ”xis the change in displacement or position andtis the time.

06

The given data

- The efficiency for conversion of electrical power by the motor is:ฮท=95.0%.

- The potential difference across the batteries:ฮ”V=12.0V.

- The mass of the electric car is:m=750kg.

07

(a) To accelerate the calculation of current

The initial speed of the car is:vi=0.

The final speed of the car is: vf=25.0m/s.

The time interval of the acceleration is:

ฮ”t=(1.00min)(60s1min)=60.0s

Given that the efficiency of the conversion is95.0%, so:

(0.95)Eelectrical=ฮ”K

Substitute Pฮ”t for Eelectrical from Equation (1) and the change in kinetic energy of the car is from Equation (3):

(0.95)Pฮ”t=12mvf2โˆ’12mvi2

Substitute Iฮ”V for P in the above equation.

(0.95)(Iฮ”V)ฮ”t=12mvf2โˆ’12mvi2

Rearrange the above equation for current as below.

I=mvf22(0.95)ฮ”Vฮ”t

Solve the above equation by substituting known numerical values in the above equation.

I=(750kg)(25.0m/ms s)22(0.95)(12.0V)(60.0s)=343A

Therefore, in this case the amount of current is 343A.

08

(b) To climb the calculation of current

The change in height of the car is: ฮ”y=2.00ร—102m.

The time interval of climbing is:

ฮ”t=(2.00min)(60s1min)=120s

The speed of the car, v=25.0m/ms s.

The force exerted by the car to overcome air resistance and friction,F=5.00ร—102N

The electrical energy Eelectricalsupplied by the battery in this case goes into increasing only the gravitational potential energy Ug of the car and exerting a positive work W to overcome air resistance and friction.

The kinetic energy of the car is constant because the car is moving with constant speed. Given efficiency of the conversion is 95.0%.So:

(0.95)Eelectrical=ฮ”Ug+W

Substitute Iฮ”Vฮ”t for Eelectical and mgฮ”y for ฮ”Ug, and Fd for W in the above equation.

(0.95)Iฮ”Vฮ”t=mgฮ”y+Fd

Here, F is the force exerted by the car and d is the distance moved by the car.

(0.95)Iฮ”Vฮ”t=mgฮ”y+Fvฮ”t

Rearrange the above equation for current I as below.

I=mgฮ”y+Fvฮ”t(0.95)ฮ”Vฮ”t

Substitute numerical values:

I=(750kg(9.80m/ms2 s2)(2.00ร—102m)+(5.00ร—102N)(25.0m/ms s)(120s)0.95ร—12.0Vร—120s=2.17ร—103A=(2.17ร—103A)(1kA1000A)=2.17kA

Therefore, in this case the amount of current is 2.17kA.

09

(c) To travel the calculation of current:

The speed of the car is v=25.0m/ms s.

The force exerted by the car to overcome air resistance and friction is,F=5.00ร—102N.

Overcome air resistance and friction. Given that efficiency of the conversion is95.0%.

Therefore,

(0.95)Eelectrical=W

Here, W is found from Equation (5):

(0.95)Iฮ”Vฮ”t=Fd

Here, F is the force exerted by the car and d is the distance moved by the car found from Equation (6):

l(0.95)Iฮ”Vฮ”t=Fvฮ”t(0.95)Iฮ”V=Fv

Solve the above equation for current I as below.

I=Fv(0.95)ฮ”V

Substitute known numerical values in the above expression.

I=(5.00ร—102N)(25.0m/ms s)0.95ร—12.0V=1.10ร—103A=(1.10ร—103A)(1kA1000A)=1.10kA

Therefore, in this case the amount of current is 1.10kA.

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