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Coletta - Physics Fundamentals 1/e
WebAssign
Coletta - Physics Fundamentals 1/e (Homework)
Current Score : 25 / 28
James Finch
Physics - College, section 1, Fall 2019
Instructor: Dr. Friendly
Due : Monday, January 28, 2030 00:00 ESTLast Saved : n/a Saving... ()
1.1/2 points | Previous AnswersColFunPhys1 4.P.006.
The particle shown below is at rest, where F1 = 6.90 N, and F2 = 10.4 N. Find the magnitude and
direction of F.
magnitude 13.8 N
direction
16.32
° counterclockwise from the +x-axis
2.1/1 points | Previous AnswersColFunPhys1 4.P.007.
A boat is pulled at constant velocity by the two forces shown below, where F = 49.0 N, and θ =
11.0°. Find the horizontal force exerted on the boat by the water. The +x direction is to the right.
in the +x direction.
96.2 N
3.2/2 points | Previous AnswersColFunPhys1 4.P.015.
A 3.00 kg mass is acted upon by four forces in the horizontal (xy) plane, as shown in figure below,
where F1 = 36.0 N, F2 = 48.0 N, F3 = 29.0 N, and F4 = 45.0 N. Find the acceleration of the mass.
magnitude 13.8 N/kg
direction
16.6
° counterclockwise from the +x-axis
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4.7/9 points | Previous AnswersColFunPhys1 4.WE.004.
Example 4: Forces on a Man
Find the forces acting on a standing man whose mass is 104.0 kg.
Part 1 of 5 - Solution
According to Eq. 4-5, the man experiences a force w in the downward
downward
direction (the direction of g), and the magnitude of this force is
w = mg = (104.0 kg)( 386.4 inches/s^2
1020 N
9.80 m/s^2 ) = 1.024 kiloN
Part 2 of 5 - Solution
This force can also be expressed in pounds.
w = 1019.2 N
1.00 lb
4.45 N
= 229 lb
229 lb
Part 3 of 5 - Solution
Since the man is standing at rest, his acceleration is zero and so the second law implies that there
must be another force to cancel the weight and produce a resultant force equal to zero, as shown in
the figure above. This other force is produced by the contact between his feet and the surface on
which he is standing. We denote this surface force by S and use the second law to solve for it:
ΣF = ma = 0
S + w = 238.8
0
Thus
S = (No Response)
Part 4 of 5 - Solution
The equation above says that the forces are oppositely directed and have equal magnitudes:
S = w = 1020 N
1020 N
Part 5 of 5 - One Step Further
We could have just as easily solved this problem using Newton's second law in component form.
Taking the positive y-axis in the upward direction, we have
ΣFy = may = 0
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Coletta - Physics Fundamentals 1/e
Thus
$$S−w
=0
Therefore
S = w = 1020 N
1020 N
Conclusion
The forces S and w are equal here because the man is stationary. It is possible for him to increase
the force S by pushing down on the ground with a force greater than his weight. By Newton's third
law, the upward force on his feet will then be greater. There would then be a resultant upward force of
magnitude S − w, and the man would accelerate upward. In other words, by pushing on the ground
with a force greater than his weight, the man can jump.
5.2/2 points | Previous AnswersColFunPhys1 4.P.026.
The block in the figure below rests on a frictionless surface. Find its instantaneous acceleration when
the spring on the left is compressed 4.43 cm while the spring on the right is stretched 10.5 cm. Each
spring has a force constant of 1.17
103 N/m.
magnitude 87.34 N/kg
direction
to the right
6.1/1 points | Previous AnswersColFunPhys1 3.A2L.002.
For which cases is the acceleration the same for the motion described in both columns?
Case
(A)
(B)
(C)
Column 1
Column 2
A car goes from 0 to 60 mps in 6 s along
A car goes from 60 to 0 mph in 6 s along a
a straight highway.
straight highway.
A race car travels around a circular track
A race car travels around the same circular
at 50 mph.
track at 100 mph.
A ball is thrown straight up. It rises 20 ft.
A ball is dropped straight down. It falls 20
Ignore the effects of the air.
ft. Ignore the effects of the air.
Case C only
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7.3/3 points | Previous AnswersColFunPhys1 4.P.029.
Find the tension in each string in the figure below, where θ1 = 42.5°, θ2 = 20.5°, and w = 28.5 N.
T1 = 29.96 N
T2 = 23.58 N
T3 = 28.5 N
8.3/3 points | Previous AnswersColFunPhys1 4.P.031.
A crate weighing 5.61
102 N is lifted at a slow, constant speed by ropes attached to the crate at A
and B in the figure below, where θ = 38.5°. These two ropes are joined together at point C, and a
single vertical rope supports the system.
(a) Find the tension T1 in the vertical rope.
561 N
(b) Find the tensions T2 and T3 in the other ropes.
T2 = 451 N
T3 = 451 N
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9.2/2 points | Previous AnswersColFunPhys1 4.P.039.
Two blocks connected by a string are on a horizontal frictionless surface. The blocks are connected to
a hanging weight by means of a string that passes over a pulley as shown in the figure below, where
m1 = 1.90 kg, m2 = 2.75 kg, and m3 = 5.05 kg.
(a) Find the tension T in the string connecting the two blocks on the horizontal surface.
9.69 N
(b) How much time is required for the hanging weight to fall 10.0 cm if it starts from rest?
0.2 s
10.
1/1 points | Previous AnswersColFunPhys1 4.P.044.
A car is stuck in a mudhole. In order to move the car, the driver attaches one end of a rope to the car
and the other end to a tree 11.5 m away, stretching the rope as much as possible as shown in the
figure below. The driver then applies a horizontal force of 3.60
102 N perpendicular to the rope at
its midpoint. The rope stretches, with its center point moving 51.0 cm to the side as a result of the
applied force. The car begins to move slowly. What is the tension in the rope?
2043 N
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11.
2/2 points | Previous AnswersColFunPhys1 19.CQ.008.
A lemon with a copper rod and a steel paperclip stuck in the sides acts as a battery (see figure).
(a) What is the source of the emf?
the paperclip
the voltmeter
the rod
the lemon
(b) This battery is not at all practical as a source of energy because of its very high internal
resistance. What would happen to the voltmeter's reading if the lemon were connected to a 1
Ω resistor to form a complete circuit? (Assume the resistor is connected at the same points as
the copper rod and steel paperclip.)
The reading would decrease, but not to zero.
The reading would not change.
The reading would go to zero.
The reading would increase.
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