Problem#1
A high diver of mass 70.0 kg jumps off a board 10.0 m above the water. If his downward motion is stopped 2.00 s after he enters the water, what average upward force did the water exert on him?
Answer:
We find the diver’s impact speed by analyzing his free-fall motion:
vf2 = vi2 + 2ax
vf2 = 0 + 2(–9.80 m/s2)(–10 m)
vf = –14.0 m/s
Now for the 2.00 s of stopping, we have
vf = vi + at
0 = –14 m/s + a(2.00 s)
a = 7.00 m/s2
Call the force exerted by the water on the diver F. Using
∑F = ma
+F – (70.0 kg)(9.80 m/s2) = (70.0 kg)(7.00 m/s2)
F = 1.18 kN
Problem#2
A crate of weight Fg is pushed by a force P on a horizontal floor. (a) If the coefficient of static friction is µs and P is directed at angle θ below the horizontal, show that the minimum value of P that will move the crate is given by
P = µsFg sec θ/(1 - µs tan θ)
(b) Find the minimum value of P that can produce motion when µs = 0.400, Fg = 100 N, and θ = 0°, 15.0°, 30.0°, 45.0°, and 60.0°.
Answer:
(a) The crate is in equilibrium, just before it starts to move. Let the normal force acting on it be n and
the friction force, fs.
Resolving vertically:
∑Fy = 0
n – P sin θ – mg = 0
n = P sin θ + mg
Horizontally:
∑Fx = 0
P cos θ – fs = 0
P cos θ = fs
But,
fs ≤ µsn
i.e.,
fs ≤ µs(P sin θ + mg), or
P cos θ ≤ µs(P sin θ + mg)
P(cos θ - µs sin θ) ≤ µsmg
Divide by cos θ :
P(1 - µs tan θ) ≤ µsmg sec θ
Then
Pmin = µsmg sec θ/(1 - µs tan θ)
(b) Pmin = (0.400)(100 N) sec θ/[1 – (0.400)tan θ]
Pmin = (40.0 N) sec θ/[1 – (0.400)tan θ]
| θ | 0.00 | 15.0 | 30.0 | 45.0 | 60.0 |
| P(N) | 40.0 | 46.4 | 60.1 | 94.3 | 260 |
If the angle were 68.2° or more, the expression for P would go to infinity and motion would become impossible.
Problem#3
Review problem. A block of mass m = 2.00 kg is released from rest at h = 0.500 m above the surface of a table, at the top of a θ = 30.0° incline as shown in Figure 2. The frictionless incline is fixed on a table of height H = 2.00 m. (a) Determine the acceleration of the block as it slides down the incline. (b) What is the velocity of the block as it leaves the incline? (c) How far from the table will the block hit the floor? (d) How much time has elapsed between when the block is released and when it hits the floor? (e) Does the mass of the block affect any of the above calculations?
 |
| Fig.2 |
Answer:
(a) Following the in-chapter Example about a block on a frictionless incline, we have
∑F = ma
mg sin θ = ma
a = g sin θ
a = (9.80 m/s2) sin 30.00 = 4.90 m/s2
(b) The block slides distance x on the incline, with
Sin 30.00 = 0.50 m/x
x = 0.50 m/sin 30.00 = 1.00 m
so that
vf2 = vi2 + 2ax
vf2 = 0 + 2(4.90 m/s2)(1.00 m)
vf = 3.13 m/s
after time is
xf = xi + vit + ½ at2
1.00 m = 0 + 0 + ½ (4.90 m/s2)t2
ts = 0.639 s
(c) Now in free fall, we use
yf = yi + vyit + ½ gt2
with vyi = vi sin 3300 = (3.13 m/s) sin 3300 = –0.733 m/s
and vxi = vi cos 3300 = (3.13 m/s) cos 3300 = 2.71 m/s
then
0 = 2.00 m + (–0.733 m/s)t + ½ (–9.80 m/s2)t2
4.90t2 + 1.56t – 2.00 = 0
t = [–1.56 m/s ± {(1.56 m/s)2 – 4(4.9 m/s2)(–2.00 m)}1/2]/[2(4.90 m/s2)]
t = 0.499 s
so that
xf = vxit = (2.71 m/s)(0.499 s) = 1.35 m
(d) total time T = t + ts = 0.499 s + 0.639 s = 1.14 s
(e) The mass of the block makes no difference.
Problem#4
A 1.30-kg toaster is not plugged in. The coefficient of static friction between the toaster and a horizontal countertop is 0.350. To make the toaster start moving, you carelessly pull on its electric cord. (a) For the cord tension to be as small as possible, you should pull at what angle above the horizontal? (b) With this angle, how large must the tension be?
Answer:
With motion impending,
∑Fy = 0
n + T sin θ – mg = 0
n = mg – T sin θ
then
fs = µsn = µs(mg – T sin θ)
and
∑Fx = 0
T cos θ - fs = 0
T cos θ = µs(mg – T sin θ)
T(µssin θ + cos θ) = µsmg
T = µSmg/(µssin θ + cos θ)
(a) To minimize T, we maximize µssin θ + cos θ
d(µssin θ + cos θ)/dt = 0
µscos θ – sin θ = 0
so, tan θ = µs = 0.350
then, θ = tan-1(0.350) = 19.30
(b) With this angle, θ = 19.30, then
T = (0.350)(1.30 kg)(9.80 m/s2)/(0.350 sin 19.30 + cos 19.30)
T = 4.21 N
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