An unbalanced force that influences the motion of an object will cause it to experience an acceleration ( a ). If such an object is opposed by some equal and opposite force ( F ), it will move with a constant velocity ( v ). Maintenance of this type of motion requires energy ( J ) in the form of work:
W = Fd
Furthermore, this force type is non-varying. A springy system is different in that increasing increments of force are needed to compress a spring by some distance ( s ). This increase is linear, and its graph is the slope of a straight line:
f ( x ) = mx + b
Fs = ks
As a spring is compressed ( or extended ), the energy requirements needed to do so increase. Since work is a product of a force acting over some distance, the area ( A ) under each of the slopes above corresponds to the energy that is needed to compress a spring, thus increasing its potential energy ( PE ). Each area is that of a triangle of dimensions summed up as follows:
A = ½ bh
In regard to each graph, the base ( b ) refers to x-axis values of ( s ), and the height of each line is denoted by ( h ). Substitution of appropriate values into the area-of-a-triangle equation yield the following results:
ΔPE = ½ ( s )( Fs )
ΔPE = ½ ( s )( ks )
ΔPE = ½ ks2
From another vantage point, the spring’s PE is summed up as a average of the force applied to the system to attain a displacement:
Favg = ½ ( Ff + Fi )
Favg = ½ ( Ff + 0 )
Favg = ½ ks
And,
W = ( Favg )( s )
W = ½ ks2 = ΔPE
Q: Four people with masses of 70 kg enter a car with a mass of 1,100 kg. Each shock absorber ( spring ) compresses by a distance of 2.5 cm ( 2.54 cm = 1 inch ). What is the elastic constant ( k ) of each spring? How much potential energy is stored within each spring upon being compressed? What is the equivalent elastic constant for the system as a single unit? Finally, where does the energy come from?
A: In order for increases in internal energy to occur in each spring, they must be compressed. For this reason, the system is considered to be in equilibrium prior to the entry of the passengers and driver ( Soon to be all passengers 🙂 ). This makes the mass of the car irrelevant to our calculations. Since the net force on the shock absorbers is due to the net mass of the people inside the car, the net force imparted ( a.k.a. “ weight “ ) is determined in the following manner:
70 kg x 4 = 280 kg
Fw = mg
Note: Please recall that the gravitational constant of acceleration ( g ) near the surface of the earth is 9.8 m/s2.
Fw = ( 280 kg )( 9.8 m/s2 )
Fw = 2,744 N
The task of stabilizing the weight is shared by the shock absorbers. For this reason, the upward force that each of them impart on the system is ¼ Fw :
Fs = ¼ ( 2,744 N )
Fs = 686 N
Fs = ks
686 N = ( k )( 0.025 m )
k = 2.7 x 104 N / m
Since the system is in equilibrium, the net force acting upward on the car is equal to the car’s weight:
Fs = k’s
Fs = Fw
2,744 N = ( k’ )( 0.025 m )
k’ = 1.1 x 105 N / m
This is approximated four times the spring constant ( k ) of each spring:
( k’ / k ) = [ ( 1.1 x 105 N / m ) / ( 2.7 x 104 N / m ) ] ≅ 4.1 times larger
We obtain the elastic potential energy stored in each spring via substitution into the elastic ( PE ) equation:
ΔPE = ½ ks2
ΔPE = ½ ( 2.7 x 104 N / m )( 0.025 m )2
ΔPE = 8.44 J
The entire system contains four times this quantity of PE:
ΔPEtotal = 4 x 8.44 J = 33.8 J
Finally, where did this energy come from? Well, the initial PE of the car is due to it being raised by some height ( h ) above the earth’s surface:
PE = mgh
PE = ( 1,100 kg )( 9.8 m/s2 )( some distance greater than the length of the shock absorbers )
As the car moves closer to the earth’s surface, ( h ) decreases, and so does the car’s PE. Some of this loss of energy is gained by the shock absorbers, and some of it is lost as thermal energy ( heat ).
RENAISSANCE PHYSICS 