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NAME ______________________________________ DATE _______________ CLASS ____________________
Holt Physics
Problem 4D
FINAL VELOCITY AFTER ANY DISPLACEMENT
PROBLEM
A bicyclist riding in the rain suddenly applies the brakes and slides to a
stop. If the acceleration is −9.5 m/s2, what is the coefficient of kinetic friction between the bicycle’s rubber tires and the wet concrete?
SOLUTION
1. DEFINE
Given:
anet = −9.5 m/s2
g = 9.81 m/s2
Unknown:
2. PLAN
mk = ?
Choose the equation(s) or situation: Use Newton’s second law to describe the
forces acting on the bicycle.
Fnet = m anet = −Fk
Use the definition of frictional force to express Fk in terms of the coefficient of
friction.
Fk = mk Fn = mk (mg)
Rearrange the equation(s) to isolate the unknown(s):
m anet = −mkmg
a et
mk = − n
g
3. CALCULATE
Substitute the values into the equation(s) and solve:
Copyright © by Holt, Rinehart and Winston. All rights reserved.
−(−9.5 m/s2)
mk = 
9.81 m/s2
mk = 0.97
4. EVALUATE
The coefficient of static friction for rubber and most surfaces is high. This is indicated by the value for rubber and wet concrete. Even under these conditions, ms is
nearly 1.
ADDITIONAL PRACTICE
1. Blocks of ice are slid down a metal chute with an incline of 12.0° above
the horizontal. The blocks undergo a constant acceleration of 1.22 m/s2.
What is the coefficient of kinetic friction between the ice and the chute?
2. A force of 1760 N is required to start moving a bundle of wooden
planks up a ramp. If the ramp’s incline is 17° and the mass of the
planks is 266 kg, what is the coefficient of static friction between the
planks and the ramp?
Problem 4D
Ch. 4–7
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3. A bundle of bricks is pulled up a ramp to a construction site. The bundle has a mass of 5.1 × 102 kg, and the incline of the ramp is 14°. If the
minimum force needed to move the bricks up the ramp is 4.1 × 103 N,
what is the coefficient of static friction between the bricks and the
ramp?
4. A force of 5.00 N to the left causes a 1.35 kg book to have a net acceleration of 0.76 m/s2 to the left. What is the frictional force acting on the
book?
5. A jar is slid horizontally across a smooth table. If the coefficient of kinetic friction between the jar and the table is 0.20, what is the magnitude of the jar’s acceleration?
6. A skier is pulled by an applied force of 2.50 × 102 N up a slope with an
incline of 18.0°. If the combined mass of the skier and skis is 65.0 kg
and the net acceleration uphill is 0.44 m/s2, what is the frictional force
between the skis and the snow?
7. If the skier in problem 6 skis down the same hill, what will the skier’s
acceleration be?
9. A horse must exert a force of 590 N just to keep a sleigh from sliding
down a snowcovered hill. The component of the sleigh’s weight down
the slope of the hill is 950 N, and the coefficient of static friction between the sleigh’s runners and the snow is 0.095. What is the normal
force exerted by the ground on the sleigh? What is the sleigh’s mass if
the slope of the hill is 14.0°?
10. A freight elevator accelerates upward at 1.20 m/s2. A crate is lifted inside the elevator. In order to move the crate along the floor of the elevator, a worker must exert a force of 1.50 × 103 N at an angle of 10.0°
below the horizontal on the upper corner of the crate. If the coefficient
of static friction is 0.650, what is the normal force that the elevator
floor exerts on the crate? What is the crate’s mass?
Ch. 4–8
Holt Physics Problem Bank
Copyright © by Holt, Rinehart and Winston. All rights reserved.
8. A crate is pushed across a level floor by a force of 3.00 × 102 N exerted
at an angle of 20.0° below the horizontal. The coefficient of kinetic
friction between the crate and floor is 0.250. If the crate’s velocity is
constant, what is the magnitude of the normal force exerted on the
crate by the floor? What is the mass of the crate?
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Additional Practice 4D
Givens
Solutions
1. anet = 1.22 m/s2
Fnet = m anet = mg(sin q) − Fk
q = 12.0°
Fk = mk Fn = mk mg(cos q)
g = 9.81 m/s2
m anet + mk mg(cos q) = mg(sin q)
2.04 m/s2 − 1.22 m/s2
(9.81 m/s2)(sin 12.0°) − 1.22 m/s2
g(sin q ) − anet
mk = 
= 
= 
2
(9.81 m/s )(cos 12.0°)
(9.81 m/s2)(cos 12.0°)
g (cos q )
0.82 m/s2
= 0.085
mk = 
(9.81 m/s2)(cos 12.0°)
2. Fapplied = 1760 N
Fnet = Fapplied − mg(sin q) − Fs,max = 0
q = 17.0°
Fs,max = ms Fn = ms mg(cos q)
m = 266 kg
ms mg(cos q ) = Fapplied − mg(sin q)
g = 9.81 m/s2
2
Fapplied − mg(sin q) 1760 − (266 kg)(9.81 m/s )(sin 17°)
ms =  = 
2
(266 kg)(9.81 m/s )(cos 17°)
mg(cos q )
1760 − 760 N
1.00 × 103 N
ms = 
= 
2
(266 kg)(9.81 m/s )(cos 17°)
(266 kg)(9.81 m/s2)(cos 17°)
ms = 0.40
Fnet = Fapplied − mg(sin q) − Fs,max = 0
3. m = 5.1 × 102 kg
Fs,max = ms Fn = ms mg(cos q)
q = 14°
3
Fapplied = 4.1 × 10 N
2
g = 9.81 m/s
ms mg(cos q ) = Fapplied − mg(sin q)
4.1 × 103 N − (5.1 × 102 kg)(9.81 m/s2)(sin 14°)
Fapplied − mg(sin q)
ms =  = 
(5.1 × 102 kg)(9.81 m/s2)(cos 14°)
mg(cos q)
4.1 × 103 N − 1.2 × 103 N
2.9 × 103 N
= 
ms = 
2
2
2
(5.1 × 10 kg)(9.81 m/s )(cos 14°)
(5.1 × 10 kg)(9.81 m/s2)(cos 14°)
4. Fapplied = 5.0 N to the left
Fnet = m anet = Fapplied − Fk
Fk = Fapplied − m anet
m = 1.35 kg
2
anet = 0.76 m/s to the left
Fk = 5.0 N − (1.35 kg)(0.76 m/s2) = 5.0 N − 1.0 N = 4.0 N
Fk = 4.0 N to the right
Fnet = m anet = Fk
5. mk = 0.20
2
g = 9.81 m/s
Fk = mkFh = mkmg
m mg
anet = k  = mk g = (0.20)(9.81 m/s2)
m
anet = 2.0 m/s2
V
V Ch. 4–6
Holt Physics Solution Manual
Copyright © by Holt, Rinehart and Winston. All rights reserved.
ms = 0.60
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Givens
Solutions
6. Fapplied = 2.50 × 102 N
m = 65.0 kg
Fnet = m anet = Fapplied − mg(sin q) − Fk
Fk = Fapplied − mg(sin q) − manet
q = 18.0°
Fk = 2.50 × 102 N − (65.0 kg)(9.81 m/s2)(sin 18.0°) − (65.0 kg)(0.44 m/s2)
2
anet = 0.44 m/s
Fk = 2.50 × 102 N − 197 N − 29 N = 24 N = 24 N downhill
Fnet = m anet = mg(sin q) − Fk
7. m = 65.0 kg
2
24 N
F
anet = g(sin q) − k = (9.81 m/s2)(sin 18.0°) −   = 3.03 m/s2 − 0.37 m/s2 = 2.66 m/s2
65.0 kg
m
g = 9.81 m/s
Fk = 24 N
anet = 2.66 m/s2 downhill
q = 18.0°
8. Fapplied = 3.00 × 102 N
Fx,net = Fapplied(cos q) − Fk = 0
q = 20.0°
Fy,net = Fn − mg + Fapplied(sin q) = 0
mk = 0.250
Fk = mkFn
2
Fapplied(cos q) (3.00 × 10 N)[cos(−20.0°)]
Fn =  =  = 1130 N
0.25°
mk
1130 N + (3.00 × 102 N)[sin(−20.0°)]
Fn + Fapplied(sin q)
 = 
m= 
9.81 m/s2
g
103 0 N
1130 N − 103 N
=  2 = 105 kg
m = 
9.81 m /s
9.81 m/s2
Fnet = Fapplied + Fs,max − Fdownhill = 0
9. Fapplied = 590 N
Fdownhill = 950 N
Fs,max = ms Fn = ms mg(cos q)
ms = 0.095
ms Fn = Fdownhill − Fapplied
q = 14.0°
Fdownhill − Fapplied 950 N − 590 N 360 N
 =  =  = 3800 N
Fn = 
ms
0.095
0.095
Copyright © by Holt, Rinehart and Winston. All rights reserved.
Fn = 3800 N perpendicular to and up from the ground
Fn
3800 N
m = 
= 
= 4.0 × 102 kg
g(cos q)
(9.81 m/s2)(cos 14.0°)
10. anet = 1.20 m/s2
Fx,net = Fapplied(cos q) − Fs,max = 0
3
Fapplied = 1.50 × 10 N
Fs,max = msFn
q = −10.0°
(1.50 × 103 N)[cos(−10.0°)]
Fapplied(cos q)
Fn =  =  = 2.27 × 103 N
0.650
ms
ms = 0.650
g = 9.81 m/s2
Fn = 2.27 × 103 N, upward
Fy,net = m anet = Fn − mg + Fapplied(sin q)
m(anet + g) = Fn + Fapplied(sin q)
Fn + Fapplied(sin q) 2.27 × 103 N + (1.50 × 103 N)[sin(−10.0°)]
 = 
m= 
1.20 m/s2 + 9.81 m/s2
anet + g
2.27 × 103 N − 2.60 × 102 N 2.01 × 103 N
m = 
= 
= 183 kg
11.01 m/s2
11.01 m/s2
V
Section Five—Solution Manual
V Ch. 4–7