Download Δx = vxt 32 m = 16 m/s

First you have to find the components of the initial velocity.
vy = 20sin37o = 12 m/s
20 m/s
37o
vy
vx = 20cos37o = 16 m/s
vx
A) How long does it take to reach the fence?
32 m = 16 m/s (t) t = 2.0 s
Δx = vxt
B) How high (Δy) is it when it reaches the fence?
2
2
Δy = voyt + ½at = 12(2) + .5(-9.8)2 = 4.4 m
The ball will be 1.9 m above the fence (4.4-2.5)
C) On the axes below, sketch the horizontal and vertical
components of the velocity of the ball as functions of time
until the ball reaches the plane of the fence.
The horizontal velocity
is not affected by
gravity and is constant.
The vertical velocity is
affected by -9.8 m/s2.
It decreases approx.
10 m/s every second or
5 m/s every half second.
2A) FN = mg = (10 kg)9.8m/s2 = 100 N
Ffr = μFN = .2(100 N) = 20 N
Fnet = FA - Ffr = 50 N - 20 N = 30 N
a = 3 m/s2
30 N = 10 kg(a)
Fnet = ma
2B)
Ffr
FN
T
10 kg
FG
T
5 kg
FG
2C) The accelerating force is the same, but in this case
the force accelerates a 10 kg and 5 kg block.
a = 2 m/s2
30 N = 15 kg(a)
Fnet = ma
3A) N1 and N2 have to
provide a force equal
to the weight of the
person and the beam.
N1 + N2 = (40kg + 50kg)9.8 m/s2 = 882 N
B) N1 is no longer a force so the weight of the beam has
to produce torque to balance the weight of the girl. N2
is the axis of rotation.
5m
40 kg
Στc = Στcc
40kg(9.8 m/s2)5m = 50kg(9.8)d
x = 4.0 m (from Δ)
50 kg
Girl is 1.0 m from end X.
d
A 0.40 kg object moves in a straight
line under the action of a net force.
The graph above shows the velocity
as a function of time for the object
during a 25 s interval. At time t = 0,
the object is at the position x = 0.
Acceleration is the slope for each section.
A) On the grid below, sketch a graph of the acceleration as a
function of time for the object. Label the scale for the
acceleration.
B) Calculate the position of the
object at t = 5.0 s.
C) On which segment of the graph
is the net force acting on the
object zero?
____ A ____ B ____ C Justify your answer.
D) Calculate the net force on the object during the first 3.0 s
of the motion.
E) Calculate the amount of work done on the object by the net
force during the first 15 s of the motion.
F) For the interval t = 15 s to t = 25 s, is the work done on the
object by the net force positive, negative, or zero?
____ Positive ____ Negative ____ Zero
Justify your answer.