Physics 198
Homework 6
Due September 15
1. (a) If the kinetic energy of an arrow is doubled, by what factor has its speed increased? (b) If
its speed is doubled, by what factor does its kinetic energy increase?
Solution:
(a) Since KE  12 mv 2 , then v  2  KE  m and so v  KE . Thus if the kinetic energy is
doubled, the speed will be multiplied by a factor of
2 .
(b) Since KE  12 mv 2 , then KE  v 2 . Thus if the speed is doubled, the kinetic energy will be
multiplied by a factor of 4 .
2. A spring has a spring stiffness constant, k, of 440 N m . How much must this spring be
stretched to store 25 J of potential energy?
Solution:
The elastic potential energy is given by U  12 kx 2 where x is the distance of stretching or
compressing of the spring from its natural length.
2U
225J 
U  12 kx2  x 


x  0.34m
k
440 N / m
3. A 1200-kg car rolling on a horizontal surface has speed v  65 km h when it strikes a
horizontal coiled spring and is brought to rest in a distance of 2.2 m. What is the spring stiffness
constant of the spring?
Solution:
Assume that all of the kinetic energy of the car becomes PE of the compressed spring.
2
1
2
mv 2  12 kx 2

 1m s  
1200 kg   65 km h  

2
mv
 3.6 km h    8.1  10 4 N m

 k 2 
2
 2.2 m 
x
4. A spring with k  53 N m hangs vertically next to a ruler. The end of the spring is next to the
15-cm mark on the ruler. If a 2.5-kg mass is now attached to the end of the spring, where will the
end of the spring line up with the ruler marks?
Solution: The spring will stretch enough to hold up the mass. The force exerted by the spring
will be equal to the weight of the mass.
mg  k  x   x 
mg
k

 2.5 kg   9.80 m
53 N m
Thus the ruler reading will be 46 cm  15 cm  61cm .
s2
  0.46 m
Physics 198
Homework 6
Due September 15
5. The roller-coaster car shown in Fig. is dragged up to point 1 where it is released from rest.
Assuming no friction, calculate the speed at points 2, 3, and 4.
Solution:
Since there are no dissipative forces present, the mechanical energy of the roller coaster will be
conserved. Subscript 1 represents the coaster at point 1, etc. The height of point 2 is the zero
location for gravitational PE. We have v1  0 and y1  35 m .
Point 2:
1
2
mv12  mgy1  12 mv22  mgy2 ; y2  0  mgy1  12 mv22 

v2  2 gy1  2 9.80 m s 2
Point 3:
1
2
 35 m   26 m s
mv12  mgy1  12 mv32  mgy3 ; y3  28 m  mgy1  12 mv32  mgy3 

v3  2 g  y1  y3   2 9.80 m s 2
Point 4:
1
2
  7 m   12 m s
mv12  mgy1  12 mv42  mgy4 ; y4  15 m  mgy1  12 mv42  mgy1 

v4  2 g  y1  y4   2 9.80 m s 2
  20 m   20 m s
6. In a two-mass system shown below, each mass is 2 kg and each spring constant k=100N/m.
How many modes of longitudinal vibration does this system have? How many modes of
transverse vibration does this system have? Calculate frequency modes.
m
m
Solution:
If the masses are free to move in all three coordinate directions, 2-mass system has 2 modes of
longitudinal vibration and 4 modes of transverse vibration.
fa 
1
2
k
1

m 2
fa 
1
2
3k
1

m 2
100 N // m
 1.13Hz
2kg
3100 N // m 
 1.95 Hz
2kg
Physics 198
Homework 6
Due September 15
7. A fisherman notices that wave crests pass the bow of his anchored boat every 3.0 s. He
measures the distance between two crests to be 6.5 m. How fast are the waves traveling?
Solution:
The wave speed is given by v   f . The period is 3.0 seconds, and the wavelength is 6.5 m.
v   f   T   6.5 m   3.0 s   2.2 m s