Question

a vehicle with a mass of 800kg experience a friction force of 180N and travels at a constant speed of 54km/h. 1.Calculate the power required from the engine to travel at the constant speed. 2.The work done to accelerate the vehicle from 54km/h with an acceleration of 2 m/s for 5 seconds

Answer 1

1. Power is work per time. Work is force times distance in the direction of force. P= fd/t = f v.

I don't have anymore time, but I saw you're on so I decided to give you some information!

Answer 2

@LinuxMint still need help here?

Answer 3

hi

ive done the following, im 90% sure im on the right track although it took me a while to actually understand it.

Power = force x speed, the force is 180N the speed is how fast the car is travelling in metres per second

Power = 180 x 54000 / 3600 (the answer is in watts)

Work done is force x distance. Force to accelerate vehicle use F = ma = 800 x 2 + 180 (i need to add in the friction of 180N as this must be overcome)

F = 1780N.

Using s = ut + 1/2 at^2 you can work out how far it travels during this acceleration (just plugged the numbers in!) u =54, a = 2, t = 5

And then multiply it by the force I worked out.

what do you guys think?

Answer 4

Second part is not OK. The mass is already moving at a constant 54Km/h (because the foce of 180N is applied to compensate friction). The work done to accelerate at 2m/sec^2 for 5 secs is additional to the 180N, then you do not have to add 180N (they are already there)

Answer 5

The additional work contributes to increase Kinetic energy
Initial velocity = Vo(54 Km/h)
Final Velocity =Vo+at=Vo+10

Increment of kinetic energy = \[\Delta E_k=\frac{ 1 }{ 2 }m(v_0+10)^2-\frac{ 1 }{ 2 }mv_0^2\]

Answer 6

I get 160 KJoules
The formula is coming from:
\[dE=F·ds=m·a·ds=m·\frac{ dv }{ dt }ds=m \left( \frac{ ds }{ dt }\right)dv=m·v·dv\]Then Work done between x1 and x2 is:\[W=E=\int\limits_{x_0}^{x_1}Fds=\int\limits_{v_0}^{v_0+10}m·v·dv=\frac{ 1 }{ 2 }m \left[ v^2 \right]_{v_0}^{v_0+10}\]

Answer 7

I mean work done between xo and x1

Answer 8

Another way to do it:
Distance moved in 5 seconds
\[d=v_0t+\frac{ 1 }{ 2 }a·t^2=15·5+\frac{ 1 }{ 2 }·2·25=100\]
The additional force is: \[F=m·a=800·2=1600\]Additional work\[W=F·d=1600·100=160,000 J=160KJ\]

Answer 9

i believe you are correct regarding the issue of friction, i actually understand now how to see the effect of forces upon a vehicle with a constant speed. i think the keyword here was CONSTANT which i missed totally. i see when it comes to calculations its better to simplify before tackling a problem regarding v=54km/h=15m/s.

this is my second week studying engineering science. i only started studying after 14 years of laziness. i guess no time like the present.

thanks for helping me. much appreciated

Answer 10

you are welcome

Answer 11

hi , regarding your explanation of friction effect in the calculation of work done, im thinking about this question and i wonder wouldn't i need to overcome friction just to increase velocity because essentially i am increasing velocity due to my acceleration after i have achieved constant speed?

Answer 12

Absolutely not. Friction was already overcome, otherwise the mass would not move at a constant speed in the beginning. If the mass is moving at constant speed that means there is no acceleration and therefore no net force on it. If you now apply a force (additional) the effect of this force will be to increase kinetic energy (velocity) Read carefully the problem statement:"The work done to accelerate the vehicle from 54km/h with an acceleration of 2 m/s for 5 seconds" If the 180 Newton force was used to accelarate, then it would not be balancing friction and the initial velocity would not be 54 km/h and of course would not be constant (the mass would be decelerating due to friction)

Answer 13

There are two energies here: The force used to keep velocity constant (balancing friction) and the force to increase velocity from 15m/sec up to 25 m/sec. It is the work done by this force what we are looking for

Answer 14

thanks for the help much appreciated , i now understand the question completely and your explanation makes a lot of sense to me . just one question : do you have a website to recommend to study moments, beams, shearing force force diagrams etc. my textbook sucks

Answer 15

Which level are you studying? Is it high school or college?

Answer 16

first year college

but had studied a couple of years back so i'm quite rusted in terms of learning

Answer 17

well, I am afraid I studied Physics many years ago and by that time internet was not even invented. What I usually do is refresh things with my books (probably some of them out of print) and google. What is the book you are using?

Answer 18

engineering science by MJJ van Rensburg

Answer 19

and is covering only Physics?

Answer 20

yes

Answer 21

im studying part time , so im doing my course per module.

Answer 22

im doing civil engineering course

Answer 23

I have always liked the books by Paul Tipler. Good books for a general course in Physics (expensive, though)
Also Alonso-Finn are really impressive, advanced (and tough) for Engineering courses (expensive too)
Halliday-Resnick is a good option (some people dont like it, I do)
"An Introduction to Mechanics" by Kleppner. I do not have it but always heard a very good critique though it is a bit old fashioned format (was used in MIT)

Answer 24

my current work is in waste water management, and i really enjoy it , pretty exciting stuff. im in the process of advancing my career hence civil . i will check out some of the books recommended. ive also heard good critique about kleppner