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.

Unizor - Physics4Teens - Mechanics - Power

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



Definition of Mechanical Power



To analyze the motion, we often use a concept of speed.

Let's assume that an object moves in some inertial reference frame, and
the distance covered by it from its initial position along its
trajectory is a function of time S(t).

Recall the definition of speed of an object as the rate, at which this object covers the distance along its motion along a trajectory.

In case of a uniform motion we can simply divide the distance S, covered during time t, by the time t to get the speed:

V = S/t

In case of non-uniform motion the speed changes and at any particular moment of time t an instantaneous speed can be calculated using differentials:

V(t) = dS(t)/dt



To analyze the mechanical work performed to achieve certain results, we often use a concept of power.

Let's assume that something or someone performs certain work and, as the time goes by, the work performed is a function of time W(t).

The power is the rate, at which the work is performed.

If during the time t the work performed is W, we define the average power of the whoever or whatever performs the work as

P = W/t

Most likely, at equal in length but different time intervals the amount
of work performed will be different. For example, when a car starts, its
engine should give a car an acceleration, which requires more work per
unit of time than to maintain a constant speed on a smooth straight
road.

In cases like this we can talk about an instantaneous power as a function of time that can be calculated using differentials:

P(t) = dW(t)/dt



Consider an example of an object in uniform motion against the force of friction with a constant speed V.

The force F that moves it forward must be equal in
magnitude and opposite in direction to the force of friction to maintain
the constant speed. Since the friction is constant, the force F must be constant as well.

The distance S covered as a function of time t is

S(t) = V·t

Therefore, the work performed by the force F during the time t is

W(t) = F·S(t) = F·V·t

From the definition of power follows that the power this force F exhorts is

P(t) = dW(t)/dt = F·V

As an example, the car engine exhorts the same power and consumes the
same amount of gas per unit of time, if the car uniformly moves along a
straight road. This power is used to generate a force sufficient to
overcome the friction of wheels and air resistance.



Consider a more general case, when the motion is not uniform.

Assume, an object of mass m moves as a result of action of force F(t), where t is time. The distance it covers is S(t).

Then during an infinitesimal time interval from t to t+dt the work performed by this force will be

dW(t) = F(t)·dS(t)

Considering the Newton's second law,

F(t) = m·a(t),

where a(t) is acceleration.

Increment of distance is

dS(t) = V(t)·dt,

where V(t) is an object's speed.

Also, by definition of acceleration,

a(t) = dV(t)/dt

Therefore,

dW(t) = (m·dV(t)/dt)·V(t)·dt =

= m·V·dV(t)

Power exhausted by this force is, therefore

P(t) = dW(t)/dt =

= m·V·dV(t)/dt =

= m·V(t)·a(t)



From the definition of power as amount of work per unit of time or, more precisely, the first derivative of work by time

P(t) = dW(t)/dt

follows that the unit of measurement of power is joule/sec called watt.

Expanding the definition of joule as newton·meter,

1 watt = 1J/sec = 1N·m/sec

Obvious extensions of unit of power watt are

kilowatt = 1,000 watt and

megawatt = 1,000,000 watt.



There is an old unit of power called horsepower.

Metric horsepower, derived from lifting up against a force of gravity on Earth a weight of mass 75 kg with a constant speed of 1 m/sec, is related to watt unit as

1(metric HP) =

=75(kg)·9.8(m/sec²)·1(m/sec)≅

≅ 735.5(W)

For historical reasons there is also a mechanical horsepower, defined as 33,000 pound-feet per minute, related to watt unit as

1(mechanical HP) ≅ 745.7(W)



So, a car engine of 200 mechanic horsepower has the power of about 149,140 watt.



Watt, as a unit of measurement, was called in honor of James
Watt, an 18th century Scottish scientist who was one of the first to
research a concept of power, developed steam engines and measured the
power of a horse.



Now let's address the concept of power in a case of rotation with constant angular speed. An example is lifting a bucket of water from a well.

Assume, a bucket of water has a mass m and we lift it with constant linear speed V with an angular speed of the well's wheel ω=V/R, where R - radius of a well's wheel.



Since the speed is constant, the force F that acts on a bucket equals to m·g, where g - acceleration of the free fall.

At the same time, if the wheel is turned by some motor and R is the radius of its shaft, the motor manufacturer provides technical characteristic not only of the power, but also of a torque of a motor.

Remember that the torque equals to

τ = F·R



So, on one hand, we have expressed the power of a motor P in terms of unknown force F and linear speed of a bucket:

P = F·V = F·R·ω

On another hand, we expressed the torque of this motor in terms of the same unknown force F and a radius of its shaft:

τ = F·R

Substituting torque τ for F·R in the formula for power, we can find the relationship between the power of a motor and is torque:

P = τ·ω



Notice the similarity between the formula for power in case of uniform motion along a straight line P=F·V and formula for power in case of rotation with constant angular speed P=τ·ω. Instead of force F in case of straight line motion, we use torque τ for rotation and, instead of linear speed V for straight line motion, we use angular speed ω for rotation.



Let's check this with real data about a particular engine.

Below is a graph representing the power and torque of Ford Motor Company 6.7L Power Stroke diesel V-8.



As you see, the power and torque grow relatively monotonically until
some engine limitations start playing significant role. While in the
area of monotonic growth, we can take a particular angular speed, say,
1400 RPM (revolutions per minute) and see that the power of an engine
equals approximately 174 HP (mechanical horsepower) and the torque is
about 650 Lb-Ft (pound-feet).

Let's check if the relationship between power and torque derived above is held in this case.

First of all, we transform all units into standard physical measures defined in SI:

1 RPM (revolutions per minutes) =

= 2π radian per minute =

= 2π/60 radians/sec =

= 0.1048 rad/sec

1 HP = 745.7 W = 745.7 J/sec

1 Lb-Ft = 1.35582 J

Substituting all the above, we see that

angular speed is equal to

1400·0.1048 = 146.6 rad/sec

power is

174·745.7 = 129,751.8 J/sec

torque equals to

650·1.35582 = 881.3 J



Now we can check the relationship between the power, torque and angular speed

P = τ·ω

Indeed,

881.3·146.6 = 129,198.6

As we see, the difference between power and a product of torque by
angular speed is minimal, attributable to imprecise measurement of the
parameters and graph reading.

Unizor - Physics4Teens - Mechanics - Work - Problems

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



Problems on Mechanical Work



Problem A



An object of mass m is pushed up a plane inclined to horizon at angle φ with constant acceleration a to a height h against a friction with coefficient of kinetic friction μ.

(a) What is the amount of work necessary to achieve this goal?



(b) Consider the following real conditions of this experiment:

m = 10kg(kilograms)

g = 9.8m/sec²(meters per sec²)

h = 0.5m(meters)

φ = 30°(degrees)

μ = 0.2

a = 0.1m/sec²(meters per sec²)

The work W=F·S is measured in N·m=kg·m²/sec² units, called joule and abbreviated as J.

Calculate the work W performed in this experiments in joules.



Solution

Let F be the force that pushes the object up the slope, P=m·g is the object's weight, R is the force of resistance from the friction, S is the distance this object moves to reach the height h.

(a) F−P·sin(φ)−R = m·a

R = P·cos(φ)·μ

W = F·S

S = h/sin(φ)

F = P·sin(φ) + R + m·a =

= m·[g·sin(φ) + g·cos(φ)·μ + a]

W = F·h/sin(φ) =

=m·h[g+g·cot(φ)·μ+a/sin(φ)]

(b)
W = 67J(joules)





Problem B



An object of mass m is pushed by a constant force F down a slope of a plane inclined to horizon at angle φ. Initially, it's at rest, the final speed is V. There is a friction with coefficient of kinetic friction μ.

(a) What is the time t from the beginning to the end of the object's motion?

(b) What is the distance S this object moved until reaching the final speed V?

(c) What is the amount of work W performed by this force?

(d) Consider the following real conditions of this experiment:

F = 20N(newtons)

m = 10kg(kilograms)

g = 9.8m/sec²(meters per sec²)

V = 0.5m/sec(meters per sec)

φ = 5°(degrees)

μ = 0.7

Calculate the work W performed in this experiments in joules, distance S and time t of motion.



Solution

P=m·g is the object's weight,

R = P·cos(φ)·μ is the force of resistance from the friction,

S is the distance this object moves to reach the speed V,

a is the acceleration of this object.

Then from the Newton's Second Law

F+P·sin(φ)−R = m·a

(we added P·sin(φ) because an object moves down a slope)

Now we can find an acceleration of our object:

F+m·g·sin(φ)−m·g·cos(φ)·μ =

= m·a

a = F/m + g·sin(φ) − g·cos(φ)·μ

Knowing acceleration a and the correspondence between acceleration, final speed and time V=a·t, we can determine time:

(a) t = V/a

Now we can find the distance

(b) S = a·t²/2 = V·t/2

The work performed by the force F equals to

(c) W = F·S

(d) a = 0.066(m/sec²)

t = 7.576(sec)

S = 1.894(m)

W = 37.879(joules))





Problem C



An object of mass m is dropped down from a certain height above a surface of some planet. At the moment it hits the ground its speed is V.

What is the work W performed by the force of gravity?



Solution

Let the acceleration of free falling on this planet is a, the time of falling t and the height S.

Then

F = m·a

V = a·t

t = V/a

S = a·t²/2 = V²/(2·a)

W = F·S = m·V²/2

Notice, the work depends only on mass and final speed - the results of
action, not on the height, nor acceleration of free falling, nor on
time.