Dynamic Engineering Systems

Dynamic Engineering Systems 1
DYNAMIC ENGINEERING SYSTEMS
A Research Paper on Engineering Science By
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Dynamic Engineering Systems 2
A horizontal force of 80 N acts on a mass of 6 kg resting on a horizontal surface. The mass is
initially at rest and covers a distance of 5 m in 0.92 s under the action of the force. Assuming
there are no energy losses due to air resistance and therefore that the acceleration is constant:
(a) Calculate the total energy expended in the acceleration.
Force applied, F = 80N
Mass of box, m = 6kg
Distance covered, d = 5m
Time, t = 0.92s
Initial velocity of the box, u = 0m/s
Total energy expenditure = Work done x Distance
E = F x d
E = 80 x 5
= 400J
(b) Plot a graph of the kinetic energy of the mass against time.
Since acceleration, a = am/s2
Equation of motion; d = ut + 1
2 at2
Will be reduced to: 5 = 0 + 1
2 a (0.92)2
Therefore, a = 11.81m/s2
Velocity, v = u + at

Dynamic Engineering Systems 3
V = 0 + (11.81)t
= 11.81t
Therefore, KE of box = 1
2 MV2
= 1
2x 6 x (11.81t)2
= 418.43t2
The values of KE and t can be:
Time KE
0 0
0.2 19.73
0.4 66.94
0.6 150.63
0.8 267.79
0.92
(c) Plot a graph of the kinetic energy of the mass against distance.

Dynamic Engineering Systems 4
Let s be distance covered after time t
V2 – U2 = 2as
V2 – (0)2 = 2 x 11.81 x s
V2 = 23.62S
KE = 1
2MV2
= 1
2 x 6 x 23.62 x S
= 70.86s
Table of KE and s
Distances (s) KE
0 0
1 70.86
2 141.72
3 212.58
4 283.44
5 354.3
(d) Calculate the coefficient of friction between the mass and the surface.