Name______________________________
Newton Laws Review Sheet
1. Draw a complete vector diagram for the incline plane problem, find all the forces
and the acceleration listed below if mass of the object is 4.0 kg, coefficient of
friction is .25 and the angle of incline is 40o
Fg = mg
Fg = 4.0(9.8) = 39.2N
FN = 30.0N
Fg paral = 39.2 sin 40 = 25.2
Fg perpen = 39.2 cos 40 = 30.0
Fg = -39.2N
Fg perpendicular = Fn
Fg par=-25.2N
Ff = µ Fn
Ff = µ Fg perpendicular
Fg per= -30N
Ff = .25 (30) = 7.5N
Ff = 7.5N
Fnet = Fg paral + (- Ff)
Fnet = 25.5 – 7.5 = 17.7
Fnet = =17.7 N
Fnet = ma
a = 4.4 m/s2
17.7 = 4.0 (a)
a = 4.4 m/s2
2. Find the acceleration of a 2.0 kg object if the coefficient of friction is .20 and the
force exerted on it is 16N. Complete a free-body diagram.
m = 2.0 kg
µ = .20
Fn
Ff
Fa
Fa = 16 N
a=?
Fnet = Fa + (-Ff)
Ff = µ Fn
ma = Fa - Ff
Ff = µ Fg
2.0(a)= 16-3.9
Ff = µ mg
a = 6.0 m/s2
Fg
Ff = .20(2.0)(9.8)=3.9
3. The famous ice skating duo, Hans and Gretchen are preparing for practice when nasty
Hans gives Gretchen a push of 200 Newton. If Gretchen has a mass of 60 kg and Hans is
90 kg, find each of their accelerations. Include a force diagram.
H
Fn = Fg
Fa = 200N
F = ma
200 = (90)a
G
Fn = Fg
Fa = -200
-200 = (60)a
a = 2.22 m/s2
Fg = (90)(9.8)
F = ma
a = 3.33 m/s2
Fg = (60)(9.8)
4. If you use a horizontal force of 30.0N to slide a 12.0 kg wooden crate across a floor at
a constant velocity, what is the coefficient of friction between the crate and the floor?
Fa = 30.0N
Fn
m = 12.0 kg Ff
Fa
µ=?
Fg
Fnet = Fa + (-Ff)
Ff = µ Fn
0 = 30 - Ff
Ff = µ Fg
30 = Ff
Ff = µ mg
30 = µ(12)(9.8)
µ = .26
5. A Junior physicist is standing on a spring scale in an elevator to see if the teacher was
correct about elevator rides. If the student has a mass of 75 kg, calculate what the scale
will read if: a) the elevator moves at constant velocity b) the elevator accelerates upward
at 1.5 m/s2. Show a free-body diagram
a)
Fn
Fnet = Fn + (-Fg)
Fn
b)
ma = Fn - mg
0 = Fn – mg
Fg
Fnet = Fn + (-Fg)
mg + ma = Fn
mg = Fn
(75)(9.8) + (75)(1.5) = Fn
(75)(9.8) = Fn
Fn = 847.5 N
735N = Fn
Fg
6. A sled with a mass of 50 kg is pulled along flat, snow-covered ground. That static
friction coefficient is .30, and the kinetic friction coefficient is .10.
a) What does the sled weigh?
Fn
Ff
Fnet = Fn + (-Fg)
Fa
Fg
0 = Fn - Fg
mg = Fn
(50) (9.8) = Fn
490 N = Fn
b) What force will be needed to start the sled moving? (we are starting at rest so
we use coefficient of static friction)
Fn
Ff = µ Fn
Ff = µ Fg
Ff
Fa
Ff = µ mg
Ff = .3(50)(9.8)
Fg
Ff =147N
c) What force is needed to keep the sled moving at a constant velocity?
Fn
Ff
Ff = µ Fn
Fa
Ff = µ Fg
Ff = µ Fg
Fg
Ff = (µ) m g
Ff = .1(50)(9.8)
Ff = 49N
d) Once moving, what total force must be applied to the sled to accelerate it at
3.0 m/s2?
Fn
Fnet = Fa + (-Ff)
ma = Fa -Ff
Ff
Fa
Fg
50(3) = Fa - 49
Fa = 199 N