A 14 kg block sits on a frictionless ramp compressing a spring 2 m. The spring constant,
k, of the spring is 150 N/m. The spring is released and the block slides up the ramp, as
pictured below.
V=0

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:

which three statements are true about all alloys?

massxgravity x heght=weightxheight= pe

4900=490xh

h=10