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									PHY131                                    Ch 6 to 10 Exam                      Name:
Answer 3 of the below 4; clearly mark out the one your skip. One of the below can receive bonus points .


6.17                                                                           10.16       A massless spring hangs from the ceiling. A
You push downward on a trunk at an angle 25° below                             10 kg bike hangs from the spring. How much does its
the horizontal with a force of 750 N. If the trunk is                          potential energy increase, if another 20 kg bike is also
on a flat surface and the coefficient of static friction                       attached to it? The spring constant is 4000 N/m.
is 0.66, what is the most massive trunk you will be able                       FW = mg         mg = kx0               mg = kxf
to move?                                                                       FHookes = -kx   100 = 4000 x0          300 = 4000 x0
 84 kg      73 kg 66 kg 101 kg                                                                 x0 = 1/40 m            x0 =3/40 m

   FPush-x   -        Ff      = mT a (just over 0 to move)                     ∆Ug = ½   k    ( xf2   -   x 02 )
cos25°(750)) - 0.66 (mg + sin25°(750)) = 0                                                          2
                                                                               ∆Ug = ½(4000) ((3/40) – (1/40)2)
m = 72.7 kg                                                                    ∆Ug = 10 Joules

10.19     A block starts from rest at the top of a 40°                         9.17
inclined plane and encounters a spring, of constant 2.4                        A car heading north collides at an intersection with a
kN/m, rigidly attached to the plane. If the block's                            truck of equal mass heading east. If they lock
mass is 34 kg and it compresses the spring by 32 cm,                           together and travel at 28 m/s at 35° north of east
find the distance the block traveled before it                                 just after the collision, how fast was the car initially
encountered the spring.                                                        traveling?
                                                                                16 m/s         64 m/s           24 m/s          32 m/s
m g      h         = ½kx2
m g (0.32+d)sin40° = ½kx2                                                      m vy + 0 = (2m) sin 35° 28m/s
d = 24.2                                                                       vy = 32.1 m/s
Spring compression is the key to this problem, this is not a stiff spring,
and must include in problem.




Ch7            What is the acceleration of the system?                       (6(g) – 8(g) sin30°) –         Ff         =     mT a
                  (μ is equal on both surfaces)                              (60 –       40) – [0.1 (6(g) + 8(g)cos30°] = (6+6+8) a
                                                                             a = 0.35 m/s2




9.3        A stationary 2.68 kg object is struck by a stick. The object                                  F    dt           = ∆mv
experiences a horizontal force given by F = at - bt3 , where t is the time in                         1500∫t dt – 20∫t3 dt = ∆mv
milliseconds from the instant the stick first contacts the object. If a = 1500                        ∆v = 1.996 m/s
N/ms and b = 20 N/ms3, what is the speed of the object just after it comes
away from the stick (t = 2.74 ms)?
2.3 m/s          16 m/s         2 m/s           14 m/s
8.6             The figure shows two wires that are tied to a 1000 g mass which
revolves in a horizontal circle at a constant speed of 8.0 m/s. What is the tension in
the upper wire?

 FT-up-y     =      FT-down-y + m g
sin30° FT-up = cos30°FT-down + 10 N
FT-up = FT-down + 20 N

 FT-up-x     +    FT-down-x = mv2 / r
cos30° FT-up + cos30°FT-down = 1 82 / 0.866

FT-down = 32.7 N                      FT-up = 52.7 N




PHY131                                Ch 6 to 10 Exam                       Name:
Answer 3 of the below 4; clearly mark out the one your skip. One of the below can receive bonus points .


6.12                                                                       10.41
Kieran takes off down a 50 m high, 10° slope on his                        A 50.0 gram ice cube can slide without friction up and
jet-powered skis. The skis have a thrust of 410 N.                         down a 30.0 degree slope. The ice cube is pressed
The combined mass of skis and Kieran is 50 kg (the                         against a spring at the bottom of the slope,
fuel mass is negligible). Kieran's speed at the bottom                     compressing the spring 10.0 cm. The spring constant is
is 40 m/s. What is the coefficient of kinetic friction                     20.0 N/m. When the ice cube is released, what
of his skis on snow?                                                       distance will it travel up the slope before reversing
 0.74    0.87    0.90     0.67                                             direction?
                                                                               Wspring = Δ m g h
d = 50/sin10°                                                              ½ k x2       = m g (sin30)(d + 0.1)
                                                                           ½(20)(0.1)2 = .05(10)(sin30)(d + 0.1)
mg h + F d        –      Ff    d = ½ m v2                                  d = 0.3 m
m(g) 50 + 410(d) - μ(cos10°)mg(d) = ½ m 402
μ = 0.734                                                                  Spring compression is the key to this problem. 10cm is significant.



10.22                                                                      10.3
A fuel tank explodes, ripping the container into two                       A prankster drops a water balloon from the top of a
pieces of masses 16 kg and 30 kg and sending them                          building on an unsuspecting person on the sidewalk
10.8 m and 7.5 m high, respectively. What was the                          below. If the balloon is traveling at 31.5 m/s when it
energy released by the explosion if all energy becomes                     strikes a person's head (1.5 m above the ground), how
the KE of the fragments?                                                   tall is the building? Neglect air resistance.
 400 kJ         400 J           3.9 kJ     3900 kJ
                                                                            54 m 52 m 51 m 55 m
ΔK = m1 g h1 + m2 g h2                                                      m g (h – 1.5) = ½ m (31.5)2
ΔK = 16(10)10.8 + 30(10)7.5 =                                              h = 52.1 meters
ΔK = 3980 Joules
Ch7   What is the acceleration of the             (8(g) sin30° - 4g sin60°) –       Ff          = mT a
system? (μ is equal on both surfaces)             ( 40 – 34.6) – [0.1 (8(g)cos30° +4(g)cos60°)] = (8+4) a
                                                          5.4         - 8.9   = 12 a
                                                  a = 0 m/s2




9.3    A stationary 2.14 kg object is struck by a stick. The                          F    dt           = ∆mv
object experiences a horizontal force given by F = at - bt3                        1500∫t dt – 20∫t3 dt = ∆mv
, where t is the time in milliseconds from the instant the           (a) 17 m/s    ∆v = 2.500 m/s
stick first contacts the object. If a = 1500 N/ms and b =            (b) 2.9 m/s
20 N/ms3, what is the speed of the object just after it              (c) 20 m/s
comes away from the stick (t = 2.74 ms)?                             (d) 2.5 m/s




8.4             A 100 g bead on a 60 cm long string is swung in a vertical circle
about a point 200 cm above the floor. The tension in the string when the bead is at
the very bottom of the circle is 6.0 N. A very sharp knife is suddenly inserted, as
shown in the figure, to cut the string directly below the point of support. How far
to the right of the center of the circle does the ball hit the floor?



 2 / 2      3/3               8 / 8     8 / 8                  4/4
               2                          2
    d     =½gt          Fc = FT – mg = m v / r        vx = Δx / Δt
(2 – 0.6) = 5 t2        6 – 1 = 0.1 (vx)2 / 0.6       5.48 = Δx / 0.53
t = 0.53 sec            vx = 5.48 m/s                 Δx = 2.90 meters

								
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