Unformatted text preview: Newton's second law applied in the vertical direction gives µ sFN – mg = 0 or
µ s = (mg)/FN = 0.323 _____________________________________________________________________________________________ 61. SSM WWW REASONING If the effects of gravity are not ignored in Example 5, the
plane will make an angle θ with the vertical as shown in figure A below. The figure B
shows the forces that act on the plane, and figure C shows the horizontal and vertical
components of these forces. Chapter 5 Problems 279 T cos θ
L θ L T sin θ θ mg mg r A B C From figure C we see that the resultant force in the horizontal direction is the horizontal
component of the tension in the guideline and provides the centripetal force. Therefore, T sin θ = mv 2
r From figure A, the radius r is related to the length L of the guideline by r = L sinθ ;
T sin θ =
L sin θ The resultant force in the vertical direction is zero: T cosθ − mg = 0 , so that T cosθ = mg (2) From equation (2) we have
cos θ (3) Equation (3) contains two unknown, T and θ . First we will solve equations (1) and (3)
simultaneously to determine the value(s) of the angle θ . Once θ is known, we can calculate
the tension using equation (3).
SOLUTION Substituting equation (3) into equation (1): Fmg I sin θ =
cos mv 2
L sin θ Thus, sin 2 θ
Using the fact that cos2 θ + sin2 θ = 1, equation (4) can be written (4) 280 DYNAMICS OF UNIFORM CIRCULAR MOTION 1 – cos 2 θ
– cos θ =
This can be put in the form of an equation that is quadratic in cos θ. Multiplying both sides
by cos θ and rearranging yields: cos 2 θ + v2
cos θ – 1 = 0
gL (5) Equation (5) is of the form
ax 2 + bx + c = 0 (6) with x = cos θ, a = 1, b = v2/(gL), and c = –1. The solution to equation (6) is found from the
quadratic formula: x= – b ± b 2 − 4 ac
2a When v = 19.0 m/s, b = 2.17. The positive root from the quadratic formula gives x = cos θ =
0.391. Substitution into equation (3) yields
(0.900 kg)(9.80 m / s 2 )
0.391 23 N When v = 38.0 m/s, b = 8.67. The positive root from the quadratic formula gives x = cos θ =
0.114. Substitution into equation (3) yields
(0.900 kg)(9.80 m / s 2 )
0.114 77 N _____________________________________________________________________________________________ CHAPTER 6 WORK AND ENERGY
ANSWERS TO FOCUS ON CONCEPTS QUESTIONS 1. (e) When the force is perpendicular to the displacement, as in C, there is no work. When the
force points in the same direction as the displacement, as in B, the maximum work is done.
When the force points at an angle with respect to the displacement but has a component in
the direction of the displacement, as in A, the work has a value between zero and the
2. (b) Work is positive when the force has a component in the direction of the displacement.
The force shown has a component along the −x and along the +y axis. Therefore,
displacements in these two directions involve positive work.
3. (c) The work is given by W = (F cos θ)s, which is zero when F = 0 N, s = 0 m, or θ = 90º.
4. 78 kg·m2/s2
5. (a) The kinetic energy is KE = 1 mv2 . Since the velocity components are perpendicular, the
Pythagorean theorem indicates that v 2 = vEast + vNorth . Therefore, ( ) ( 3.00 kg ) ( 5.00 m/s ) + (8.00 m/s ) 2
KE = 1 m vEast + vNorth =
2 2 2 6. 115 m/s
7. (e) The work-energy theorem states that W = KEf – KE0. Since work is done, the kinetic
energy changes. Since kinetic energy is KE = 1 mv2 , the speed v must also change. Since
the instantaneous speed is the magnitude of the instantaneous velocity, the velocity must also
8. (d) The work-energy theorem indicates that when the net force acting on the particle does
negative work, the kinetic energy decreases. Since each force does negative work, the work
done by the net force must be negative, and the kinetic energy must decrease. But when the
kinetic energy decreases, the speed must also decrease, since the kinetic energy is
proportional to the square of the speed. Since it is stated that the speed increases, this answer
is not possible.
9. (b) The work-energy theorem states that the net work done on the particle is equal to the
change in the particle’s kinetic energy. However, the speed does not change. Therefore, the
kinetic energy does not change, because kinetic energy is proportional to the square of the
speed. According to the work-energy theorem, the net work is zero, which will be the case if
W1 = − W2. 282 WORK AND ENERGY 10. (d) Since the block starts from rest, the work energy theorem is
W = 1 mvf2 − 1 mv0 =
2 ( 2.70 kg ) vf2 The final speed can be obtained from this expression, since the work done by the net force is
W = (75.0 N) (cos 38.0º) (6.50 m) + (54.0 N) (cos 0.0º) (6.50 m) + (93.0 N) (cos 65.0º) (6.50 m) 11. (c) The gravitational force is a conservative force, and a conservative force does no net
work on an object moving around a closed path.
12. (e) Since air resistance always o...
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