Answer:
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 sF N
F fr = µ kF N
F = ma
F N = 930 N
µ s = .12
F fr =µ kF 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 kF 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 kF 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:
x = ?????
v i = 27
v f = 0
a = -8.33 m/s/s (slowing down)
t = don't care
Use vf 2 = vi 2 + 2ax:
0 2 = (27) 2 + 2(-8.33 m/s/s)s
x = 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 kF N
1.2 N = µk (7.84 N)
µ k = .15306 = .15
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Explanation:
Note that is not an answer that is guide
thanks po
hope it help ❤️
