HW5Sol - Physics 112 HW#5 Solutions 5-42 When ball's speed...

Physics 112 - HW #5 Solutions
5-42 When ball's speed v = v
t
/2 = mg/D /2 (half its terminal speed), f
d
= Dv
2
= mg/4.
(a) Going up:
mg
f
d
m
Σ
F = m a
( ↓
+
): mg + f
d
= ma
= mg +
mg
4
=
5mg
4
⇒ a =
5
4
g
↓
(b) Going down:
mg
f
d
m
Σ
F = m a
( ↓
+
): mg - f
d
= ma
= mg -
mg
4
=
3mg
4
⇒ a =
3
4
g
↓
5-47 The free-body diagram shows the forces acting on the plane
while turning, as viewed from the front or rear. The plane's
acceleration is a = v
2
/R in the +x direction ( ← ).
v = 240 km/h





 10
3
m
1 km




1 h
3600 s
= 66.7 m/s
Write Newton's 2nd Law,
Σ
F = m a :
+x: F
lift
sin θ = ma
x
= m





 v
2
R
(1)
+y: F
lift
cos θ - mg = ma
y
= 0 ⇒ F
lift
cos θ = mg (2)
Divide equation (1) by equation (2) to get:
F
lift
sin θ
F
lift
cos θ
= tan θ =
v
2
Rg
=
(66.7 m/s)
2
(1200 m)(9.8 m/s
2
)
= 0.378 ⇒ θ = 21°
5-75 (a) Draw a free-body diagram for all forces acting on the crate, including the friction
force f. We don't yet know whether f is static or kinetic. Write out
Σ
F = m a :
y: N - mg = ma
y
= 0 ⇒ N = mg
x: f = ma
x
= ma
If f is static friction, then f = ma = f
s
≤ µ
s
N = µ
s
mg , so that . . .