Homework 5 (327253)

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Due: Tue Oct 25 2005 11:13 PM EDT

Description
Chapter 7

 

 

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  1. Walker2 7.P.001. [236090] A farmhand pushes a 20 kg bale of hay 2.4 m across the floor of a barn. If she exerts a horizontal force of 89.0 N on the hay, how much work has she done?

Work = Force times displacement  = (89.0 N) (2.4 m) = 213.6 Nm = 214 J
[214] J

 

 

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  1. Walker2 7.P.004. [236092] The coefficient of kinetic friction between a suitcase and the floor is 0.29. If the suitcase has a mass of 70.0 kg, how far can it be pushed across the level floor with 670 J of work?

F(kinetic Friction) = mN.

N-mg = ma(vertical) = 0

N=mg

F(kinetic Friction) = m mg = (0.29) (70.0 kg) (9.81 m/s2) = 199.1 N

Work = Fd

d = Work / F = (670 J) /199.1 N =  3.36 J/N
[3.36] m

 

 

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3. Walker2 7.P.005. [236093] You pick up a 3.7 kg can of paint from the ground and lift it to a height of 1.8 m.

(a)   How much work do you do on the can of paint?

W= Fh

F = force to lift at constant velocity (accel = 0)

F-mg = 0.

W = mgh = (3.7 kg) (9.81 m/ s2) (1.8 m) = 65.3 kg m2/s2
[65.3] J
(b) You hold the can stationary for half a minute, waiting for a friend on a ladder to take it. How much work do you do during this time?

W = F d, d=0 (no CHANGE in position).
[0] J
(c) Your friend decides against the paint, so you lower it back to the ground. How much work do you do on the can as you lower it?

W = Fy (yf- yi) = mg(0-h) = -mgh = -65.3 J

Force holding can is upwards, can moves downwards, Work is negative:

You don’t have to push down to lower the can.
[-65.3] J

 

 

 

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  1. Walker2 7.P.009. [236095] A 55 kg packing crate is pulled with constant speed across a rough floor with a rope that is at an angle of 34.0° above the horizontal. If the tension in the rope is 125 N, how much work is done on the crate to move it 5.5 m?

Work = F d cos(q) = (125 N) (5.5 m) cos(34.0°) = 570 Nm
[570] J

 

 

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4.      Walker2 7.P.012. [236097] Water skiers often ride to one side of the center line of a boat, as shown in Figure 7-14. In this case, the ski boat is traveling at 15 m/s and the tension in the rope is 65 N. If the boat does 3500 J of work on the skier in 55.5 m, what is the angle between the tow rope and the center line of the boat?
Work = F d cos( )
cos( ) = Work / (F d) = (3500 J) / (65 N * 55.5 m) = 0.9702
 = 14.02°
[14]°

 
Figure 7-14

 

 

 

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6. Walker2 7.P.017. [236100] A 9.70 g bullet has a speed of 1.15 km/s.

(a) What is its kinetic energy in joules?
K = (1/2) m v2 = (0.5) ( 0.0097 kg) (1,150 m/s)2= 6414 J
[6.41] kJ
(b) What is the bullet's kinetic energy if its speed is halved?
Divide by four.
[1.6] kJ
(c) What is the bullet's kinetic energy if its speed is doubled?
Multiply by four.
[25.7] kJ

 

 

 

 

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7. Walker2 7.P.019. [236101] A 0.21 kg pinecone falls 15 m to the ground, where it lands with a speed of 16 m/s.

(a) How much work was done on the pine cone by air resistance?
Ei = m g h + 0 = (0.21 kg) (9.81 m/s2) (15 m) = 30.9 J
Ef = 0 + (1/2) m v2 = (1/2) (0.21 kg) (16 m/s)2= 26.88 J
Ei + Work done by air friction = Ef
Work = Ef - E i = (26.88 J) - (30.9 J) = - 4.02 J
[-4.02] J
(b) What was the average force of air resistance exerted on the pine cone?
<F> d = Work
<F> = Work / d = (-4.02 J) / (-15 m) = + 0.268 N
[0.268] N (upward)

 

 

 

 

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8. Walker2 7.P.025. [236105] A spring with a force constant of 2.7 104 N/m is initially at its equilibrium length.

(a)   How much work must you do to stretch the spring 0.056 m?

Force changes with stretch, cannot use W=Fd (must integrate F(x)dx)

Work done be conservative force = -(change in potential energy)

Force applied to stretch spring = - spring force

Work done by external force to stretch spring = - work done by spring

   = +(Change in potential energy) = (1/2)kx2 – 0

Work = (1/2) (2.7·104 N/m) (0.056 m)2 = 42.3 Nm
[42.3] J
(b) How much work must you do to compress it 0.056 m?

x à -x,   same work.
[42.3] J

 

 

 

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9. Walker2 7.P.027. [236106] Initially sliding with a speed of 1.7 m/s, a 1.7 kg block collides with a spring and compresses it 0.35 m before coming to rest. What is the force constant of the spring?
E = K + VSpring
Ei = (1/2) m v2 + 0 = (1/2) (1.7 kg) (1.7 m/s)2 = 2.456 J
Ef = 0 + (1/2) k x 2
Ef = Ei
k = m v2 / x 2 = (1.7 kg) (1.7 m/s)2 / (0.35 m)2 = 40.1  N/m
[40.1] N/m

 

 

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10. Walker2 7.P.028. [236107] The force shown in Figure 7-15 moves an object from x = 0 to x = 0.06 m, where the interval between vertical dashed lines is 0.02 m. How much work is done by the force?

Work = (0.6 N)(0.02 m) + (0.4 N)(0.2 m) (0.8 N) (0.2 m)
 Work = 0.036 J
[0.036] J


Figure 7-15