Unizor - Physics4Teens - Mechanics - Problems on Power

Notes to a video lecture on http://www.unizor.com



Problems on Mechanical Power



Problem A



A car engine accelerates a car of mass m from the state at rest to speed V during the time T with constant acceleration along a straight line.

Ignore loss of mass due to burning fuel.

(a) What is the power of the car engine as a function of time?

(b) If engine's power is proportional to amount of fuel supplied to it
in a unit of time (fuel burring speed), how the fuel burning speed will
change over time to assure the kind of motion described in this problem?


Solution:



(a) a = V/T

F = m·a = m·V/T

S(t) = a·t²/2 = (V/T)·t²/2

W(t) = F·S(t) = (m·V/T)·(V/T)·t²/2 = m·V²·t²/(2T²)

P(t) = dW(t)/dt = (m·V²/T²)·t



(b) The fuel burning speed is linearly increasing with time. The faster a
car goes, as it accelerates, - the more power should be supplied to
assure the constant acceleration, and the faster fuel is burning.





Problem B



A car engine supplies constant power P to a car of mass m.

Find its speed and acceleration as a function of time.

Is speed linearly increasing with time?



Solution



We will use the general formula for power as a function of mass, speed and acceleration:

P(t) = m·V(t)·a(t)

or, since a(t)=dV(t)/dt,

P(t) = m·V(t)·dV(t)/dt

Therefore,

P(t)·dt = m·V(t)·dV(t)

In our case P(t) is constant P. Therefore,

P·dt = m·V(t)·dV(t)

Integrating this by time from t=0 to t,

P·t = m·V²(t)/2

From this we can find speed V(t):

V(t) = √2P·t/m

Acceleration is

a(t) = dV(t)/dt = √P·/(2m·t)

Speed is not linearly increasing with time, it is proportional to a
square root of time, which means that acceleration is monotonically
diminishes, while speed increases.





Problem C



A car engine supplies constant power P to a car of mass m.

How long would it take for a car to reach speed V_max?



Solution



As we know,

P(t) = m·V(t)·a(t)

Since power P(t)=P - is constant, and acceleration is the first derivative of speed by time, this can be written as

P = m·V(t)·dV(t)/dt

Therefore,

P·dt = m·V(t)·dV(t)

Integrating this by time from t=0 to t,

P·t = m·V²(t)/2

From this we can find time t as a function of speed:

t = m·V²(t)/(2P)

Therefore, the time at the moment the speed is equal to V_max equals to

t_max = m·V²_max /(2P)

Obviously, the more powerful an engine is - the shorter will be the time interval it takes to achieve needed speed.



As an example, consider a car 2012 Tesla Model S 85 kWh .

It has a mass of 2108 kg and its engine develops the power of 310 Kw.

According to the formula above, the time it takes for this car to reach the speed of 60 miles/hour (26.8 m/sec) equal to

2108·26.8²/(2·310000) = 2.44(sec)

So, in theory, this car is capable to reach the speed of 60 miles/hour in just under 2.5 sec.