1) A car is moving with a speed of 20 m / s when the brakes are applied. The wheels lock
(stop spinning). After travelling 40 m, the car stops. Determine the coefficient of kinetic
friction between the tires and the road.
vix = 20 m / s
FN
Δdx = 40 m
vfx = 0
𝜇k = ?
2
= vix2 + 2ax Δdx
v fx
0 = (20) 2 + 2ax (40)
Fk
( a y = 0)
(m = ? )
-(20) 2 = 80a x
-(20) 2
80
Fg
ΣFy = may
ΣFx = max
F N - F g = m(0)
-F k = max
FN - Fg = 0
∴ FN = Fg = mg
mass 'm' was not needed as it
cancelled out in our equations
= ax = - 5 m / s 2
(F k = 𝜇 k F N )
-𝜇 k F N = max
max
ma x
-a x
or
∴ 𝜇k =
=
-F N
-mg
g
𝜇k =
- -5 m / s 2
9.8 m / s 2
= 0.5
2) A hockey puck slides with an initial speed of 50.0 m / s on a large frozen lake. The
coefficient of kinetic friction between the puck and the ice is 0.030. Determine the speed
of the puck after 10.0 s.
vix = 50.0 m / s
FN
𝜇k = 0.030
vfx = ?
t = 10.0 s
forces first, then kinematic equations
- need a x to calculate v fx
ΣFy = may
Ff
F N - F g = m(0)
FN - Fg = 0
( a y = 0)
(m = ? )
ΣFx = max
-F f = max
-𝜇F N = max
-𝜇 F N
-𝜇(mg)
= ax =
m
m
∴ ax = - 𝜇g
a x = - (0.030) 9.8 m / s 2
a x = - 0.294 m / s 2
Fg
∴ FN = Fg = mg
v fx = v ix + a x t
vfx = (50 m / s) + -0.294 m / s 2 (10.0 s)
∴ vfx = 47 m / s
3) You are trying to slide a sofa across a horizontal floor. The mass of the sofa is
2.0 × 10 2 kg, and you need to exert a force of 3.5 × 10 2 N to make it just begin to move.
a) Calculate the coefficient of static friction between the floor and the sofa.
b) After it starts moving, the sofa reaches a speed of 2.0 m / s after 5.0 s. Calculate the
coefficient of kinetic friction between the sofa and the floor
FN
a)
Fa
Ff
Fg
∴
( a y = 0)
( a x = 0)
ΣFy = may = 0
ΣFx = max = 0
FN - Fg = 0
Fa - Fs = 0
F N = F g = mg
Fa = Fs = 𝜇s FN
Fa
FN
= 𝜇s =
𝜇s =
Fa
mg
3.5 × 10 2
2.0 × 10 2 (9.8)
= 0.18
b) Now we have an acceleration in the x-direction to calculate and use in the ΣF x formula
with F k instead of static friction F s
vix = 0
v fx = v ix + a x t
vfx = 2.0 m / s
ax =
t = 5.0 s
ax = ?
ay = 0
ax =
𝜇k = ?
∴
v fx - v ix
t
2-0
5
F a - max
FN
𝜇k =
ΣFx = max = 0
F a - F k = max
F a - max = F k
= 0.4 m / s 2
= 𝜇k =
F a - max = 𝜇 k F N
F a - max
mg
3.5 × 10 2 - 2.0 × 10 2 (0.4)
2
2.0 × 10 (9.8)
= 0.14
4) A crate is placed on an adjustable, inclined board. The coefficient of static friction
between the crate and the board is 0.29.
a) Calculate the value of θ at which the crate just begins to slip.
b) Determine the acceleration of the crate down the incline at this angle when the coefficient o
kinetic friction is 0.26.
a) "just begins to slip" implies ΣF x = 0, and ΣF y = 0 because no motion in y-direction
FN
+y
Ff
+x
𝜃
mg
𝜃
mgcos𝜃
ΣFy = may = 0
in𝜃
mgs
ΣFx = max = 0
F N - mgcos𝜃 = 0
mgsin𝜃 - F f = 0
1 FN = mgcos𝜃
mgsin𝜃 = F f
(Ff = 𝜇FN )
mgsin𝜃 = 𝜇 s F N
2 mgsin𝜃 = 𝜇s mgcos𝜃
to isolate for 𝜃 we need to divide both sides of 2 by cos𝜃 and use the fact that
2
mgsin𝜃
cos𝜃
=
𝜇s mgcos𝜃
sin𝜃
cos𝜃
= tan𝜃
cos𝜃
mgtan𝜃 = 𝜇 s mg
divide out mg from both sides
tan𝜃 = 𝜇 s
∴ 𝜃 = tan -1 𝜇s = tan -1 (0.29) = 16.17° ≈ 16°
b) Using the same formulas for ΣF x & ΣF y as above, except a x ≠ 0
1 FN = mgcos𝜃
2 ΣFx = max
mgsin𝜃 - F k = max
mgsin𝜃 - 𝜇 k F N = max
mgsin𝜃 - 𝜇 k (mgcos𝜃) = max
∴ gsin𝜃 - 𝜇k gcos𝜃 = ax
or
divide out m from all terms
g(sin𝜃 - 𝜇 k cos𝜃) = a x
(9.8)(sin16° - 0.26cos16°) = ax = 0.252 m / s 2 [down incline]
if you don't use more digits for 𝜃 here (from your answer in part a), you won't get the
answer in the back of the book (0.28 m / s 2 )
(9.8)(sin16.17° - 0.26cos16.17°) = ax = 0.28 m / s 2 [down incline]
6) Two blocks are connected by a massless string that
passes over a frictionless pulley, as shown. The
coefficient of static friction between m 1 and the table
is 0.45. The coefficient of kinetic friction is 0.35.
Mass m 1 is 45 kg, and m 2 is 12 kg.
a) Is this system in static equilibrium? Explain.
b) Determine the tension in the string. (120 N)
c) A mass of 20.0 kg is added to m 2 . Calculate the acceleration.
a) It will only be in static equilibrium if the force of static friction on mass 1 is greater than or
equal to the tension pulling it to the right
T
+y
ΣFy = m2 ay
m 2 g - T = m 2 (0)
m2
1 m2 g = T
∴ T = (12)(9.8) = 117.6 N
F g2
+x
Ff
FN
m1
T
ΣFy = may
Ff = 𝜇s FN
FN - m1 g = 0
Ff = 𝜇s m1 g
FN = m1 g
F f = (0.45)(45)(9.8)
F f = 198.45 N
( a y = 0)
F g1
Therefore the system is in static
equilibrium because F f > T
b) From formula 1 we have T = 117.6 N ≈ 120 N
c) Now we can use m 2 = 12 kg + 20.0 kg = 32 kg. Refer to our FBDs for the following:
(Note: the whole system accelerates at the same rate, which we'll call 'a')
Mass 1
Mass 2
ΣFx = m1 ax
T - Ff = m1 a
T - 𝜇k FN = m1 a
T - 𝜇k m1 g = m1 a
T = 𝜇k m1 g + m1 a
ΣFy = m2 ay
m2 g - T = m2 a
m 2 g - (𝜇 k m 1 g + m 1 a) = m 2 a
m2 g - 𝜇k m1 g - m1 a = m2 a
m2 g - 𝜇k m1 g = m1 a + m2 a
m 2 g - 𝜇 k m 1 g = (m 1 + m 2 )a
∴
m2 g - 𝜇k m1 g
m1 + m2
=a
a=
m2 g - 𝜇k m1 g
m1 + m2
=
(32)(9.8) - (0.35)(45)(9.8)
= 2.1 m / s 2
32 + 45
7) A block of rubber is placed on an adjustable inclined plane and released from rest. The
angle of the incline is gradually increased.
a) The block does not move until the incline makes an angle of 42° to the horizontal.
Calculate the coefficient of static friction.
b) The block stops accelerating when the incline is at an angle of 35° to the horizontal.
Determine the coefficient of kinetic friction.
FN
+y
Ff
+x
𝜃
mg
𝜃
mgcos𝜃
in𝜃
mgs
a) The block starts to move once the force pulling down the plane is greater than or equal
to the force keeping it stationary, F f .
ΣFy = may = 0
ΣFx = max = 0
F N - mgcos𝜃 = 0
mgsin𝜃 - F f = 0
F N = mgcos𝜃
mgsin𝜃 = F f
mgsin𝜃 = 𝜇 s F N
mgsin𝜃 = 𝜇 s mgcos𝜃
mgsin𝜃
mgcos𝜃
= 𝜇s = tan𝜃
divide both sides
by mgcos𝜃
sin𝜃
cos𝜃
= tan𝜃
∴ 𝜇s = tan42° = 0.90
b) Our FBDs and net force equations remain the same except we are using 𝜇 k , not 𝜇 s
∴ 𝜇k = tan𝜃
𝜇k = tan35° = 0.70
divide by mgcos𝜃