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Objective
1. Find the total force of the pushing students(F
student)
2. Find the Force of Friction (Ffriction
)3. Determine the effect of adding masses to the
truck.
Scenario 1
Push an empty truck
until it reaches
10mph (4.47 m/s)
Knowledge Base
Vo = 10 mph = 4.47 m/s
Vf =0 m/s
Time taken to reach 10 mph = 8.31 s Time taken to stop = 24.58 s
Initial Acceleration
Vo = at
4.47 = a(8.31)
a = 0.538 m/s2
Car's Deceleration
Vf = Vo + at
0 = 4.47 + a(24.58)
a = - 0.182 m/s
Knowledge Base
Mass of truck = 2678 kg
Cars decel = -0.182 m/s2
Find μf
ΣF = ma = μf N
ma = μf mg
-0.182 = μf (9.8)
μf = 0.019
Knowledge Base
Mass of truck
2678 kg
Cars Deceleration
a = -0.182 m/s2
Coefficient of Friction
μf = 0.019
Initial Acceleration
a = 0.538 m/s2
Find Fstudents
ΣF = ma = Fstudents
– Ffriction
Fstudents
= [(2678)(0.538)]+[(0.019)(2678)(9.8)
= 1939.4 N
Find Ffriction
F = μf N
= (.019)(2678)(9.8)
= - 498.6 N
Scenario 2
Pushing truck with added
masses until it reaches
10mph (4.47 m/s)
Knowledge Base
Vo = 10 mph = 4.47 m/s
Vf =0 m/s
Time taken to reach 10 mph = 10.95 s Time taken to stop = 22.58 s
Initial Acceleration
Vo = at
4.47 = a(10.95)
a = 0.408 m/s2
Car's Deceleration
Vf = Vo + at
0 = 4.47 + a(22.58)
a = - 0.198 m/s2
Knowledge Base
Mass of truck = 2990 kg
Cars decel = -0.198 m/s2
Find μf
ΣF = ma = μf N
ma = μf mg
-0.198 = μf (9.8)
μf = 0.02
Knowledge Base
Mass of truck
2990 kg
Cars Deceleration
a = -0.198 m/s2
Coefficient of Friction
μf = 0.02
Initial Acceleration
a = 0.408 m/s2
Find Fstudents
ΣF = ma = Fstudents
– Ffriction
Fstudents
= [(2990)(0.408)]+[(0.02)(2990)(9.8)
= 1805.96 N
Find Ffriction
F = μf N
= (.02)(2990)(9.8)
= - 586.04 N
Final Observations
Mass = 2678 kg => Ffriction
= -498.6 N
Mass = 2990 kg => Ffriction
= -586.04 N
According to our findings, Ffriction
is directly proportional to the mass of the object.
