Overton, Mays – Midterm 2 – Due: Mar 9 2005, 10:00 pm – Inst: Turner
1
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The due time is Central
time.
001
(part 1 of 1) 10 points
An elevator has a mass of 1100 kg and car
ries a maximum load of 777 kg
.
A constant
frictional force of 4600 N retards its motion
upward.
The acceleration of gravity is 9
.
8 m
/
s
2
.
What power must the motor deliver at a
instantaneous speed of 3
.
61 m
/
s if the elevator
is designed to provide an upward acceleration
of 1
.
16 m
/
s
2
?
Correct answer: 90870
.
6 W.
Explanation:
The motor must supply the force
~
T
that
pulls the elevator upward. Applying Newton’s
second law to the elevator gives
T

f

M g
=
M a .
Thus
T
=
M
(
a
+
g
) +
f
= (1877 kg) (1
.
16 m
/
s
2
+ 9
.
8 m
/
s
2
)
+ 4600 N
= 25171
.
9 N
.
and the required instantaneous power is
P
=
T v
= (25171
.
9 N) (3
.
61 m
/
s)
=
90870
.
6 W
.
002
(part 1 of 1) 10 points
A projectile of mass 0
.
596 kg is shot from
a cannon, at height 6
.
9 m, as shown in the
figure, with an initial velocity
v
i
having a
horizontal component of 6
.
3 m
/
s.
The projectile rises to a maximum height of
Δ
y
above the end of the cannon’s barrel and
strikes the ground a horizontal distance Δ
x
past the end of the cannon’s barrel.
The acceleration of gravity is 9
.
8 m
/
s
2
.
Δ
x
v
i
57
◦
Δ
y
6
.
9 m
Find the magnitude of the final velocity
vector when the projectile hits the ground.
Correct answer: 16
.
4025 m
/
s.
Explanation:
Let :
v
x
i
= 6
.
3 m
/
s
,
x
i
= 0 m
,
y
i
= 6
.
9 m
,
and
θ
= 57
◦
.
For the horizontal component of vector
v
i
,
we have
v
x
i
=
v
i
cos
θ
; consequently
v
i
=
v
x
i
cos
θ
=
(6
.
3 m
/
s)
cos 57
◦
= 11
.
5673 m
/
s
.
The work done by gravity depends only
on the vertical distance
y
i
, since the work
involving Δ
y
adds and subtracts;
i.e.
, cancels.
The work done by gravity is
W
=
m g y
i
.
From
Δ
K
=
1
2
m v
2
f

1
2
m v
2
i
and
the workenergy theorem Δ
K
=
W ,
the final
velocity is
v
f
=
r
2
W
m
+
v
2
i
=
r
2
m g y
i
m
+
v
2
i
=
q
2
g y
i
+
v
2
i
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Overton, Mays – Midterm 2 – Due: Mar 9 2005, 10:00 pm – Inst: Turner
2
=
h
2 (9
.
8 m
/
s
2
) (6
.
9 m)
+ (11
.
5673 m
/
s)
2
i
1
/
2
=
16
.
4025 m
/
s
.
Alternate Solution:
The initial velocity
in the vertical direction is
v
y
i
=
v
x
i
tan
θ
= (6
.
3 m
/
s) tan 57
◦
= 9
.
70115 m
/
s
,
and
the vertical velocity
v
y
top
= 0 at the top, there
fore we have
v
2
y
i
=
v
2
y
top
+ 2
g
Δ
y ,
so
Δ
y
=
s
v
2
y
i
2
g
=
s
(9
.
70115 m
/
s)
2
2 (9
.
8 m
/
s
2
)
= 4
.
80165 m
.
The magnitude of the final vertical velocity
v
y
f
is

v
y
f

=
p
2
g
[
y
i
+ Δ
y
]
=
q
2 (9
.
8 m
/
s
2
) [(6
.
9 m) + (4
.
80165 m)]
= 15
.
1444 m
/
s
,
so
v
f
=
q
v
2
y
f
+
v
2
x
i
=
q
(15
.
1444 m
/
s)
2
+ (6
.
3 m
/
s)
2
=
16
.
4025 m
/
s
.
003
(part 1 of 1) 10 points
Betty weighs 419 N
.
She is sitting on a play
ground swing seat that hangs 0
.
48 m above
the ground in its rest position.
Her initial
speed is zero and her initial height above the
ground is 1
.
69 m
.
At some later time her
speed is 1
.
23 m
/
s and her height above the
ground is 1 m
.
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 Spring '08
 Turner
 Physics, Force, Friction, Correct Answer, kg, m/s, Overton

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