MHB Work energy principle and power

[Shah 72]

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

 

[topsquark]

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

 

[Shah 72]

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

 

[skeeter]

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.

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

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

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

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

 

[Shah 72]

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!