2.1 Newton’s Second Law
7
Fig. 2.1 Mass m dropped
from elevation h on a spring
with stiffness k
The solutions by the IM relationship deliver the impact velocity with similar
calculations. We have m v(t)−m v(0) =
t
0 m g ds = m g t by IM so that v(t) = g t
since v(0) = 0. This gives x(t) = g t 2 /2 + c which becomes x(t) = g t 2 /2 since
x(0) = 0. The rest of calculations are as in the solution by the NL.
The solution by the WE relationship is slightly more efficient. We have
(1/2) m v(¯ t) 2 − (1/2) m v(0) 2 =
m g
h so that the impact velocity is v(¯ t) =
√
2 g h since v(0) = 0.
System 2–3 Consider the mass-spring system and a different system of coordinates,
the coordinate x which measures the deformation of the spring relative to its
undeformed position so that x = x + h. This new coordinate is used for
convenience. Any other coordinate can be used. The solutions by the NL and
IM momentum relationship involve lengthy calculations using tools which will be
discussed later in this chapter. On the other hand the solution by the WE relationship
is much simpler. It is based on the observations that (1) the mass velocity changes
sign at the time the spring reaches its maximum deformation δ and (2) the force
acting on the mass at an arbitrary deformation x is m g − k x , where m g is the
weight of the mass and −k x denotes the response of the spring which opposes
deformation. Accordingly, the WE relationship has the form
(1/2) m v
2
3 − (1/2) m v
2
2 =
δ
0
m g − k x
dx
,
where v 2 =
√
2 g h is the impact velocity and v 3 = 0 is the velocity at the maximum
deformation. This gives the algebraic equation −(1/2)
2 g h
= m g δ − k δ 2 /2 for
δ which has the solutions δ 1,2 = δ st ±
δ 2
st + 2 h δ st , where δ st = m g/k denotes
the static deformation under the weight of the mass. Since the solution with minus
is physically inadmissible (it suggests that the spring stretches under compression
load), we conclude that the maximum deformation of the spring is
7
Fig. 2.1 Mass m dropped
from elevation h on a spring
with stiffness k
The solutions by the IM relationship deliver the impact velocity with similar
calculations. We have m v(t)−m v(0) =
t
0 m g ds = m g t by IM so that v(t) = g t
since v(0) = 0. This gives x(t) = g t 2 /2 + c which becomes x(t) = g t 2 /2 since
x(0) = 0. The rest of calculations are as in the solution by the NL.
The solution by the WE relationship is slightly more efficient. We have
(1/2) m v(¯ t) 2 − (1/2) m v(0) 2 =
m g
h so that the impact velocity is v(¯ t) =
√
2 g h since v(0) = 0.
System 2–3 Consider the mass-spring system and a different system of coordinates,
the coordinate x which measures the deformation of the spring relative to its
undeformed position so that x = x + h. This new coordinate is used for
convenience. Any other coordinate can be used. The solutions by the NL and
IM momentum relationship involve lengthy calculations using tools which will be
discussed later in this chapter. On the other hand the solution by the WE relationship
is much simpler. It is based on the observations that (1) the mass velocity changes
sign at the time the spring reaches its maximum deformation δ and (2) the force
acting on the mass at an arbitrary deformation x is m g − k x , where m g is the
weight of the mass and −k x denotes the response of the spring which opposes
deformation. Accordingly, the WE relationship has the form
(1/2) m v
2
3 − (1/2) m v
2
2 =
δ
0
m g − k x
dx
,
where v 2 =
√
2 g h is the impact velocity and v 3 = 0 is the velocity at the maximum
deformation. This gives the algebraic equation −(1/2)
2 g h
= m g δ − k δ 2 /2 for
δ which has the solutions δ 1,2 = δ st ±
δ 2
st + 2 h δ st , where δ st = m g/k denotes
the static deformation under the weight of the mass. Since the solution with minus
is physically inadmissible (it suggests that the spring stretches under compression
load), we conclude that the maximum deformation of the spring is