UNIVERSITY OF Warwick Year 2
School of Engineering Sept 2018
ES2B0 Title of Examination: -
MECHANICS OF SOLIDS AND FLUIDS
Page 1
Questions 3-4 (Section B: Dynamics) Mark
Allocation
Marks
***************************************** Begin 3 ******************************************* *** B3 ***
Q3(a)
i. The priciple of work and energy is summarised as follows: the work done on a system is equal to the
change in the system’s kinetic energy, i.e., U = Δ K.E.
Answer variations:
A close, but not completely correct answer might be: P.E. + K.E = const (which is true for conservative
forces only)
If explained, the answer: “P.E. + K.E. – Unc= const”, is fine, where Unc is work done by non-
conservative forces (3 marks)
ii. The weight potential energy of an elevated object, in terms of its mass m, gravitational acceleration
g, and its elevation h from some reference position, is given by
P.E. = mgh
Here, h=0.4 m, g=10 ms-2 (or using 9.81is fine)
P.E.= 160 Nm (2 marks)
iii. The potential energy of a spring, in terms of its stiffness k and its extension/compression s from its
un-tensioned state, is given by
P.E. = ½ks2
The unstretched spring length is 0.3m (as deduced from information in the question). From simple
trigonometry the length of the springs at position 1 is found to be 0.5m, meaning a stretch of S=0.2m at
position 1.
The total spring P.E = 2 (½ks2) = 320 Nm (4 marks)
iv. Assuming no non-conservative forces then:
P.E. + K.E = const
There is no kinetic energy at position 1 (it is initially at rest) and there is no potential energy at position
2 (the springs are unextended and the hammer is at zero elevation), as such:
K.E.2= P.E.1=160+320 = 480 Nm
Because K.E.2=½mv2
(3 marks)
The speed v= 4.9 m/s
v.
P.E. + K.E. – Unc= const
where Unc is the work done by non-conservative forces. (2 marks)
vi.
P.E.1 =K.E.2 – Unc
K.E.2= ½mv2 = 259.2 Nm
Therefore Unc = 259.2 – 480 = – 220.8 Nm (the negative sign indicates that energy is removed from (3 marks)
the system)
School of Engineering Sept 2018
ES2B0 Title of Examination: -
MECHANICS OF SOLIDS AND FLUIDS
Page 1
Questions 3-4 (Section B: Dynamics) Mark
Allocation
Marks
***************************************** Begin 3 ******************************************* *** B3 ***
Q3(a)
i. The priciple of work and energy is summarised as follows: the work done on a system is equal to the
change in the system’s kinetic energy, i.e., U = Δ K.E.
Answer variations:
A close, but not completely correct answer might be: P.E. + K.E = const (which is true for conservative
forces only)
If explained, the answer: “P.E. + K.E. – Unc= const”, is fine, where Unc is work done by non-
conservative forces (3 marks)
ii. The weight potential energy of an elevated object, in terms of its mass m, gravitational acceleration
g, and its elevation h from some reference position, is given by
P.E. = mgh
Here, h=0.4 m, g=10 ms-2 (or using 9.81is fine)
P.E.= 160 Nm (2 marks)
iii. The potential energy of a spring, in terms of its stiffness k and its extension/compression s from its
un-tensioned state, is given by
P.E. = ½ks2
The unstretched spring length is 0.3m (as deduced from information in the question). From simple
trigonometry the length of the springs at position 1 is found to be 0.5m, meaning a stretch of S=0.2m at
position 1.
The total spring P.E = 2 (½ks2) = 320 Nm (4 marks)
iv. Assuming no non-conservative forces then:
P.E. + K.E = const
There is no kinetic energy at position 1 (it is initially at rest) and there is no potential energy at position
2 (the springs are unextended and the hammer is at zero elevation), as such:
K.E.2= P.E.1=160+320 = 480 Nm
Because K.E.2=½mv2
(3 marks)
The speed v= 4.9 m/s
v.
P.E. + K.E. – Unc= const
where Unc is the work done by non-conservative forces. (2 marks)
vi.
P.E.1 =K.E.2 – Unc
K.E.2= ½mv2 = 259.2 Nm
Therefore Unc = 259.2 – 480 = – 220.8 Nm (the negative sign indicates that energy is removed from (3 marks)
the system)
