Friction Worksheet:  1
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3  4  5  6

7  8  9  10
 11  12  13  14
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 by Chris Murray, 2001, 2004
1. What is the force of friction between a block of ice that weighs 930 N and the ground if m = .12?
F_{fr} £ µ_{s}F_{N }F_{fr} = µ_{k}F_{N }F = ma
_{}F_{N }= 930 N_{ }µ_{s }= .12_{ }F_{fr } =µ_{k}F_{N } = (930)(.12) = 111.6 N = 110 N
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2. What is the coefficient of static friction if it takes 34 N of force to move a box that weighs 67 N?
F_{N }= 67 N_{ }µ_{s }= ?_{ }F_{fr } =34 N
F_{fr} = µ_{k}F_{N }34 = µ_{k}(67)_{ }µ_{s }= .507 = .51
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3. A box takes 350 N to start moving when the coefficient of static friction is .35. What is the weight of the box?
F_{N }= ?????_{ }µ_{s }= .35_{ }F_{fr } =350 N
F_{fr} = µ_{k}F_{N }350 = (.35)_{}F_{N }F_{N }= 1000 N = 1.0 x 10^{3} N
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4. A car has a mass of 1020 Kg and has a coefficient of friction between the ground and its tires of .85. What force of friction can it exert on the ground? What is the maximum acceleration of this car? In what minimum distance could it stop from 27 m/s?
First find the normal force which is the weight in this case:
F = ma = (1020 kg)(9.80 m/s/s) = 9996 N = _{}F_{N}
Then use the Friction formula to find the frictional force with the ground:
F_{fr} = = (.85)(9996 N) = 8496.6 N = 8500 N_{ }Now we can find the acceleration. The Force of friction is what speeds up the car, so
F = ma
8496.6 N = (1020 kg)(a)
a = 8.33 m/s/s = 8.3 m/s/s
So now we need to solve a cute linear kinematics problem:s = ?????
u = 27
v = 0
a = 8.33 m/s/s (slowing down)
t = don't careUse v^{2} = u^{2 + }2as:
0^{2} = (27)^{2 + }2(8.33 m/s/s)s
s = 43.76 m = 44 m
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5. Clarice moves a 800. gram set of weights by applying a force of 1.2 N. What is the coefficient of friction?
m = 800. g = .800 kg (divide by 1000)
First find the normal force which is the weight in this case:
F = ma = (.800 kg)(9.80 m/s/s) = 7.84 N = F_{N}
Next  apply the force of friction formula:
F_{fr} = µ_{k}F_{N }
1.2 N = µ_{k}(7.84 N)
µ_{k} = .15306 = .15
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6. A car has a coefficient of friction between the ground and its tires of .85. What is the mass of the car if it takes 9620 N of force to make it slide along the ground?
First, find the normal force
F_{fr} = µ_{k}F_{N }9620 N = (.85)_{}F_{N}
_{}F_{N}= 11317.65 N
Now the Normal force is also the weight, so you can find the mass using:
F = ma
11317.65 N = m(9.80 m/s/s)
m = 1154.86 kg = 1200 kg
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7. A 5.0 Kg block has a coefficient of friction of .15 on a flat surface. What is its acceleration if you exert a force of 15 N sideways on it when it is at rest? (Find the friction force first)
First find the normal force which is the weight in this case:
F = ma = (5.0 kg)(9.80 m/s/s) = 49 N = F_{N}
Then use the Friction formula to find the frictional force with the ground:
F_{fr} = = (.15)(49 N) = 7.35 N_{ }Now we set up a Newton's Second law expression, as there are two forces acting in the horizontal direction, friction and an applied force of 15 N:
<Fnet> = ma
<15 N  7.35 N> = (5)a
a = 1.53 = 1.5 m/s/s
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8. A 10. Kg block is at rest on a level surface. It accelerates from rest to 51.2 m/s in 8 seconds when you exert a force of +115 N on it sideways. What is the acceleration of the block? What is the force of friction between the surface and the block, and what is the coefficient of friction?
Let's solve a cute linear kinematics problem first:
s = don't care
u = 0
v = 51.2 m/s
a = ????
t = 8.0 sUse v = u + at:
51.2 m/s = 0 + a(8.0 s)
a = 6.4 m/s/s
Now we set up a Newton's Second law expression, as there are two forces acting in the horizontal direction, friction and an applied force of 100 N:
<Fnet> = ma
<115 N  F> = (10)(6.4 m/s/s) = 64 N
F = 51 N = F_{fr}
Next find the normal force which is the weight in this case:
F = ma = (10. kg)(9.80 m/s/s) = 98 N = F_{N}
Then  apply the force of friction formula:
F_{fr} = µ_{k}F_{N }
51 N = µ_{k}(98 N)
µ_{k} = .5204 = .52
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9. A 120 Kg log sled accelerates at 1.4 m/s/s when a horse pulls on it. What force must the horse exert if the coefficient of friction between the ground and the sled is .28?
First find the normal force which is the weight in this case:
F = ma = (120 kg)(9.80 m/s/s) = 1176 N = F_{N}
Then use the Friction formula to find the frictional force with the ground:
F_{fr} = = (.28)(1176 N) = 329.28 N_{ }Now we set up a Newton's Second law expression, as there are two forces acting in the horizontal direction, friction and an unknown applied force:
<Fnet> = ma
<F  329.28 N> = (120 kg)(1.4 m/s/s)
F = 497.28 N = 5.0 x 10^{2} N
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10. You exert a force of 24 N sideways on an object and it accelerates from 0  12 m/s over a distance of 5.2 m. You know that the coefficient of friction between the object and the ground is .58, so what is its mass?
Let's solve a cute linear kinematics problem first:
s = 5.2 m
u = 0
v = 12 m/s
a = ??????
t = don't careUse v^{2} = u^{2 + }2as:
(12 m/s)^{2} = 0^{2 + }2a(5.2)
a = 13.8462 m/s/s
Now we set up a Newton's Second law expression, as there are two forces acting in the horizontal direction, friction and an applied force:
<Fnet> = ma
<24 N  F> = (m)(13.8462 m/s/s)
F = 24 N  (m)(13.8462 m/s/s) = Frictional force.
The normal force is going to be F = ma = m(9.8 m/s/s)
Now use the formula for friction (F_{fr} = µ_{k}F_{N}) to set this expression equal to the Frictional force
F = 24 N  (m)(13.8462 m/s/s) = Frictional force.
F = µ_{k}F_{N} = .58m(9.80 m/s/s) = m(5.684 m/s/s)
so
24 N  (m)(13.8462 m/s/s) = m(5.684 m/s/s)
24 N = (m)(13.8462 m/s/s) + m(5.684 m/s/s) = m(13.8462 m/s/s + 5.684 m/s/s) = m(19.5302m/s/s)
m = (24 N)/(19.5302m/s/s) = 1.229 = 1.2 kg
11. A 15.0 kg block of wood with a kinetic coefficient of .370 is sliding to the right at 4.45 m/s is stopped by friction and a force of 25.0 N to the left. What is its deceleration? In what distance does it stop?
We know the mass, so let's find the force of friction :
F_{fr} = µ_{k}F_{N }F_{fr} = (.370)(15.0 kg)(9.80 N/kg) = 54.39 N
As it slides along the ground in the positive direction, we have this force of friction acting in the negative direction, and the force of 25.0 N in the negative direction. Newton's second law looks like:<25.0 N  54.39 N> = (15.0 kg)a
a = 5.2927 m/s/sNow let's solve a cute linear kinematics problem :
s = ??
u = +4.45 m/s
v = 0
a = 5.2927 m/s/s
t = don't careUse v^{2} = u^{2 + }2as:
0 = (4.45 m/s)^{2 + }2(5.2927 m/s/s)s
s = 1.87 m
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12. The 1835 kg Batmobile needs to stop from a speed of 48.2 m/s. Its tires have a coefficient of friction of .93 with the road, and Batman goes to full reverse thrusters on his jet engines. What would be his acceleration if he stopped in a distance of 50.0 m? What additional stopping force does he need to do this?
Let's solve a cute linear kinematics problem first:
s = 50.0 m
u = 48.2 m/s
v = 0
a = ??????
t = don't careUse v^{2} = u^{2 + }2as:
0 = (48.2 m/s)^{2 + }2a(50.0 m)
a = 23.2324 m/s/s = 23.2 m/s/s
We know the mass, so let's find the force of friction :
F_{fr} = µ_{k}F_{N }F_{fr} = (.93)(1835 kg)(9.80 N/kg) = 16724.19 NAs the batmobile moves in the positive direction, there is friction in the negative direction, and the force of the retro thrusters on the Batmobile  we will call this force F. Now we set up a Newton's Second law:
<Fnet> = ma
<F  16724.19 N> = (1835 kg)(23.2324 m/s/s)
F = 25,907.264 N = 25,900 N  which is to the left
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13. An 89,320 kg airplane landing at 83.5 m/s must stop in a distance of 415 m. The engines can generate 313 kN of reverse thrust. What must be the deceleration of the airplane? What additional force of friction from the tires does the plane need to stop? What is the minimum coefficient of friction that must exist between the tires of the plane and the runway?
Again let's solve a cute linear kinematics problem first:
s = 415 m
u = 83.5 m/s
v = 0
a = ??????
t = don't careUse v^{2} = u^{2 + }2as:
0 = (83.5 m/s)^{2 + }2a(415 m)
a = 8.40030 m/s/s = 8.40 m/s/s
Now we set up Newton's second law  we have the force of friction acting to the left (F), and the reverse thrust of the engines to the left acting as the plane moves to the right:
<F> = ma
<F  313,000 N> = (89,320 kg)(8.40030 m/s/s)
F = 437314.9036 N = 437,000 N
And now to find the coefficient of friction:
F_{fr} = µ_{k}F_{N }437314.9036 N = µ_{k}(89320 kg)(9.80 N/kg)
µ_{k} = 0.499596616 = .500
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14. An 18.5 kg block of orthoclase chondritic basalt has a coefficient of .35 between it and the floor. It is sliding to the right at 6.72 m/s and stops in a distance of 3.93 m under the influence of friction, and another force. What is its deceleration? What is the magnitude and direction of this other force?
Dog clock on rumble knight pencil respond chair:
s = 3.93 m
u = 6.72 m/s
v = 0
a = ??????
t = don't careUse v^{2} = u^{2 + }2as:
0 = (6.72 m/s)^{2 + }2a(3.93 m)
a = 5.7453 m/s/s = 5.75 m/s/s
Since we have the mass and the coefficient of friction, we had better solve for the force of friction because you never know when that might come handy:
F_{fr} = µ_{k}F_{N }
F_{fr} = (.35)(18.5 kg)(9.8 N/kg) = 63.455 N
Now we set up Newton's second law  we have the force of friction acting to the left (F), and a MYSTERY FORCE acting in an unknown direction:
<F> = ma
<F  63.455 N> = (18.5 kg)(5.7453 m/s/s)
F = 42.83385 N = 42.8 N  which would be to the left since it is negative
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15. An 8.12 kg block of wood is moving to the right with at 4.5 m/s. There is a coefficient of kinetic friction of .21 between the wood and the floor. If no other force acts along the ground, in what distance will the block stop? If the block stops sliding in 3.15 m. What is the acceleration of the block? What other force besides friction must be applied to the block?
We know the mass, so let's find the force of friction :
F_{fr} = µ_{k}F_{N }F_{fr} = (.210)(8.12 kg)(9.80 N/kg) = 16.711 N
Now, if there is no other force acting along the plane, then we are all set. As the block slides to the right, the force of friction acts to the left. (friction always opposes the direction of sliding) Newton's second law in the horizontal direction looks like:
<F> = ma
(16.711 N) = (8.12)a
a = 2.058 m/s/s
Pablo kitten chunk disgorge fraught noble house entreat Chuck:
s = ???
u = 4.50 m/s
v = 0
a = 2.058 m/s/s
t = don't care
Use v^{2} = u^{2 + }2as:
0 = (4.50 m/s)^{2 + }2(2.058 m/s/s)s
s = 4.9198 m = 4.9 m
Let's solve a cute linear kinematics problem first:s = 3.15 m
u = 4.50 m/s
v = 0
a = ??????
t = don't careUse v^{2} = u^{2 + }2as:
0 = (4.50 m/s)^{2 + }2a(3.15 m)
a = 3.2143 m/s/s = 3.2 m/s/s
As it slides in the positive direction, there is friction in the negative direction, and an unknown force in an unknown direction  so we leave it as F. Now we set up a Newton's Second law:
<Fnet> = ma
<F  16.711 N> = (8.12 kg)(3.2143 m/s/s)
F = 9.39 N = 9.4 N  which is to the left
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