Young & Friedman 738
A 2.00kg block is pushed against a spring with negligible mass and force constant k = 400 N , compressing it 0.220 m
m. When the block is released, it moves along a frictionless, horizontal surface and then up a frictionless incline
with slope 37.0◦.
(a) What is the speed of the block as it slides along the horizontal surface after having left the spring?
The forces involved here (spring force, gravity) are conservative, so the total mechanical energy remains unchanged
throughout the motion.
KEhorizontalsliding +PEhorizontalsliding = KEcompressing spring +PEcompressing spring
1
2
mv2 +0 = 0+
1
2
k
�
Δxcompressed�2
�
v = �
Δxcompressed�
�
k
m
= (0.220) 400
2.00
= 3.11
m
s
(b) How far does the block travel up the incline before starting to slide back down?
At the top of its trajectory, the block will be momentarily at rest. In part (a) we implicitly chose the zero of gravitational potential energy to be at the horizontal surface. When the block travels a distance � up the ramp, it will be a
height � sinθ above that origin.
KEtop +PEtop = KEcompressing spring +PEcompressing spring
1 � 0+mg� sinθ = 0+ k Δxcompressed�2
2
k � � = Δxcompressed�2
2mgsinθ
400 = 2(2.00) (9.81)sin(37.0)
(0.220)
2
= 0.820m
2 Young & Friedman 751
A skier starts at the top of a very large frictionless snowball, with a very small initial speed, and skis straight
down the side. At what point does she lose contact with the snowball and fly off at a tangent? That is, at the
instant she loses contact with the snowball, what angle α does a radial line from the center of the snowball to
the skier make with the vertical?
Call the skier’s mass m and the snowball’s radius r. Choose the center of the snowball to be the zero of gravitational
potential. We can look at the velocity v as a function of the angle α and find the specific αliftoff at which the skier
departs from the snowball.
If we ignore snowski friction along with air resistance, then the one workproducing force in this problem, gravity,
is conservative. Therefore the skier’s total mechanical energy at any angle α is the same as her total mechanical
energy at the top of the snowball.
KE(α) +PE(α) = KE(α = 0) +PE(α = 0)
1
m[v (α)]2 +mgr cosα = 1
m[v (α = 0)]2 +mgr 2 2
1
m[v (α)]2 +mgr cosα ≈ mgr 2
One might be tempted to include a potential energy term arising from the normal force exerted on the skier by the
snowball. Remember that the normal force is incapable of doing work because that force is always perpendicular to
the direction of an object’s motion. There cannot be a potential energy associated with it.
The last line of the above equations will be true if the “small” initial speed v (α = 0) � √gr. In that case we can
rearrange the above equation to find the requisite centripetal force for keeping the skier along the presumably circular
path.
m[v (α)]2
= 2mg(1−cosα) r
The centripetal force (due to gravity) will be mgcosα, so the skier will remain on the snowball as long as gravity
can hold her to that path, i.e. as long as
mgcosα ≥ 2mg(1−cosα)
Any radial gravitational force beyond what is necessary for the circular motion will be balanced by the normal
force—or else the skier will sink into the snowball.
The expression for αliftoff turns out to be very simple:
3cosα ≥ 2
2
αliftoff = arccos = 48.2◦
3
It has no dependence on r, m, or even g for that matter
Explanation:
