Unformatted text preview: re to sustain the constant uphill velocity than the constant
downhill velocity, we can write
P U = P D + ∆P Chapter 6 Problems 329 where P U is the power needed to sustain the constant uphill velocity, PD is the power
needed to sustain the constant downhill velocity, and ∆ P = 47 hp . In terms of SI units, 746 W 4
∆ P = ( 47 hp ) = 3.51 × 10 W 1 hp Using Equation 6.11 ( P = F v ), the equation P U = P D + ∆P can be written as FU v = FD v + ∆P
Using the expressions for FU and FD , we have
Solving for θ, we find ( FR + mg sin θ ) v = ( FR − mg sin θ ) v + ∆ P 3.51 × 104 W
–1 = sin = 2.0° 2(1900 kg)(9.80 m/s 2 )(27 m/s) ______________________________________________________________________________ ∆P 2mgv θ = sin –1 71. SSM REASONING The area under the forceversusdisplacement graph over any
displacement interval is equal to the work done over that displacement interval.
SOLUTION
a. Since the area under the forceversusdisplacement graph from s = 0 to 0.50 m is greater
for bow 1, it follows that bow 1 requires more work to draw the bow fully . b. The work required to draw bow 1 is equal to the area of the triangular region under the
forceversusdisplacement curve for bow 1. Since the area of a triangle is equal onehalf
times the base of the triangle times the height of the triangle, we have
1
2 W1 = (0.50m)(350 N) = 88 J
For bow 2, we note that each small square under the forceversusdisplacement graph has an
area of
(0.050 m)(40.0 N) = 2.0 J We estimate that there are approximately 31.3 squares under the forceversusdisplacement
graph for bow 2; therefore, the total work done is 2.0 J (31.3 squares)
= 63 J square Therefore, the additional work required to stretch bow 1 as compared to bow 2 is 330 WORK AND ENERGY 88 J – 63 J = 25 J ______________________________________________________________________________
72. REASONING When the force varies with the displacement, the work is the area beneath
the graph of the force component F cos θ along the displacement as a function of the
magnitude s of the displacement. Here, the shape of this area is a triangle. The area of a
triangle is onehalf times the base times the height of the triangle.
SOLUTION The base of the
triangle is 1.60 m, and the
“height” is 62.0 N. Therefore,
the area, which is the work, is F cos θ
62.0 N W = 1 Base × Height
2
= 1
2 (1.60 m )( 62.0 N ) s 0 1.60 m = 49.6 J 73. REASONING AND SOLUTION
a. The work done is equal to the colored area under the graph. Therefore, the work done by
the net force on the skater from 0 to 3.0 m is Area under graph = ( 31 N − 0 N )( 3.0 m − 0 m ) = 93 J from 0 to 3.0 m b. From 3.0 m to 6.0 m, the net force component F cos θ is zero, and there is no area under
the curve. Therefore, no work is done .
c. The speed of the skater will increase from 1.5 m/s when s = 0 m to a final value vf at
s = 3.0 m . Since the force drops to zero at s = 3.0 m , the speed remains constant at the
value vf from s = 3.0 m to s = 6.0 m ; therefore, the speed at s = 6.0 m is vf . We can find
vf from the workenergy theorem (Equation 6.3). According to the workenergy theorem,
the work done on the skater is given by
1
2 1
2 2
W = mvf2 − mv0 Solving for vf , we find
2W
2(93 J)
2
2
+ v0 =
+ (1.5 m/s) = 2.3 m/s
m
65 kg
______________________________________________________________________________
vf = 74. REASONING The area under the forceversusdisplacement graph over any displacement
interval is equal to the work done over that interval. From Example 14, we know that the Chapter 6 Problems 331 total work done in drawing back the string of the bow is 60.5 J, corresponding to a total area
under the curve of 242 squares (each square represents 0.250 J of work). The percentage of
the total work done over any displacement interval is, therefore,
Work done during interval
Total work done during all intervals × 100 = Number of squares under curve during interval
Total number of squares under entire curve × 100 SOLUTION
a. We estimate that there are 130 squares under the curve from s = 0 to 0.306 m ; therefore,
the percentage of work done during the interval in question is
130 squares
× 100 = 54%
242 squares b. Similarly, we estimate that there are 112 squares under the curve in the interval from
s = 0.306 m to 0.500 m; therefore, the percentage of work done during the interval in
question is
112 squares
× 100 = 46%
242 squares
______________________________________________________________________________
75. REASONING The final speed vf of the object is related to its final kinetic energy by
KE f = 1 mvf2 (Equation 6.2). The object’s mass m is given, and we will use the workenergy
2 theorem W = KE f − KE 0 (Equation 6.3) to determine its final kinetic energy. We are helped
by the fact that the object starts from rest, and therefore has no initial kinetic energy KE0.
Thus, the net work done on the object equals its final kinetic energy: W = KE f = 1 mvf2 . In a
2
graph of F cos θ versus s, such as th...
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This note was uploaded on 04/30/2013 for the course PHYS 1100 and 2 taught by Professor Chastain during the Spring '13 term at LSU.
 Spring '13
 CHASTAIN
 Physics, The Lottery

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