Work energy principle and power

In summary, a car of mass 1200 kg accelerates up a hill inclined at 10 degrees to the horizontal. The car has an initial speed of 10 m/s and a final speed of 12 m/s after 60 s. Air resistance and friction may be ignored. The average power generated by the engine can be found using two methods. The first method involves calculating the work done by the engine and dividing it by the elapsed time. The second method involves calculating the force exerted by the engine and multiplying it by the average velocity. The correct answer is found by using the work energy theorem, taking into consideration the change in kinetic and potential energy of the car.
  • #1
Shah 72
MHB
274
0
A car of mass 1200 kg accelerates up a hill inclined at 10 degree to the horizontal. The car has initial speed of 10 m/ s and final speed of 12 m/s after 60 s. Air resistance and friction may be ignored. Find the average power generated by the engine.
The and in the textbook is 36000 W, iam not able to do. Pls help
 
Mathematics news on Phys.org
  • #2
Shah 72 said:
A car of mass 1200 kg accelerates up a hill inclined at 10 degree to the horizontal. The car has initial speed of 10 m/ s and final speed of 12 m/s after 60 s. Air resistance and friction may be ignored. Find the average power generated by the engine.
The and in the textbook is 36000 W, iam not able to do. Pls help
Again what have you been able to do? You've posted a large number of these questions and have learned nothing about how to start? How about a diagram and a statement of the work energy theorem? See what you can do, then if you still are getting an incorrect answer post what you have done and we can look it over.

-Dan
 
  • #3
topsquark said:
Again what have you been able to do? You've posted a large number of these questions and have learned nothing about how to start? How about a diagram and a statement of the work energy theorem? See what you can do, then if you still are getting an incorrect answer post what you have done and we can look it over.

-Dan
Iam getting the ans 31180W, but textbook ans is 3600W
I first calculated acceleration using v= u+at and got a= 0.033m/s^2
Then calculated F- 16000sin10=1600×0.033 and got the ans F=2831.7N
Then calculated s using v^2= u^2+2as and got s= 660.67m
Avg power= work done/ time
Iam not getting the right ans. Pls help
 
  • #4
A car of mass 1200 kg accelerates up a hill inclined at 10 degree to the horizontal. The car has initial speed of 10 m/ s and final speed of 12 m/s after 60 s. Air resistance and friction may be ignored. Find the average power generated by the engine.

Shah 72 said:
Iam getting the ans 31180W, but textbook ans is 3600W

Well, I'm not getting either of those values. First off, some basic values ...

$a = \dfrac{\Delta v}{\Delta t} = \dfrac{1}{30} \, m/s^2$

$v_{avg} = \dfrac{v_0+v_f}{2} = 11 \, m/s$

$\Delta x = v_{avg} \cdot \Delta t = 660 \, m$method 1

$P_{avg} = \dfrac{\text{total work done by the engine}}{\text{elapsed time}}$

$P = \dfrac{\Delta KE + \Delta GPE}{\Delta t}$

$\color{red} P = \dfrac{\frac{1}{2}m(v_f^2-v_0^2) + mg \Delta x \cdot \sin{\theta}}{\Delta t}$

method 2

$P_{avg} = (\text{force exerted by the engine})(\text{avg velocity}) = F_e \cdot v_{avg}$

$F_e - mg\sin{\theta} = ma \implies F_e = m(g\sin{\theta} + a)$

$\color{red} P = m(g\sin{\theta}+a) \cdot v_{avg}$
 
  • #5
skeeter said:
Well, I'm not getting either of those values. First off, some basic values ...

$a = \dfrac{\Delta v}{\Delta t} = \dfrac{1}{30} \, m/s^2$

$v_{avg} = \dfrac{v_0+v_f}{2} = 11 \, m/s$

$\Delta x = v_{avg} \cdot \Delta t = 660 \, m$method 1

$P_{avg} = \dfrac{\text{total work done by the engine}}{\text{elapsed time}}$

$P = \dfrac{\Delta KE + \Delta GPE}{\Delta t}$

$\color{red} P = \dfrac{\frac{1}{2}m(v_f^2-v_0^2) + mg \Delta x \cdot \sin{\theta}}{\Delta t}$

method 2

$P_{avg} = (\text{force exerted by the engine})(\text{avg velocity}) = F_e \cdot v_{avg}$

$F_e - mg\sin{\theta} = ma \implies F_e = m(g\sin{\theta} + a)$

$\color{red} P = m(g\sin{\theta}+a) \cdot v_{avg}$
Thank you!
 

FAQ: Work energy principle and power

What is the work energy principle?

The work energy principle states that the work done on an object is equal to the change in the object's kinetic energy. In other words, when a force is applied to an object, it will either speed up or slow down depending on the direction of the force.

How is work calculated?

Work is calculated by multiplying the force applied to an object by the distance the object moves in the direction of the force. This can be represented by the equation W = Fd, where W is work, F is force, and d is distance.

What is the unit of measurement for work?

The unit of measurement for work is the joule (J). One joule is equal to the work done when a force of one newton is applied to an object and it moves one meter in the direction of the force.

What is power?

Power is the rate at which work is done or energy is transferred. It is calculated by dividing the work done by the time it takes to do the work. The unit of measurement for power is the watt (W).

How is the work energy principle related to power?

The work energy principle and power are related because power is the rate at which work is done. Therefore, the work energy principle can be used to calculate the power of a system by dividing the work done by the time it takes to do the work.

Similar threads

Replies
2
Views
986
Replies
1
Views
618
Replies
3
Views
1K
Replies
1
Views
957
Replies
16
Views
1K
Replies
9
Views
1K
Replies
8
Views
1K
Replies
17
Views
902
Replies
2
Views
1K
Replies
9
Views
1K
Back
Top