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enrmmell
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Mechanical systems - Please help me with this question! :(
Person_Number M (kg) D (m) R (m/s2)
<431383> 2700 60 1.4
Mechanical Systems Engineering
Assignment 1
An unloaded lorry of mass M kg accelerates from rest and reaches a velocity of 30mph in 10 seconds. It then travels at constant velocity for half a minute before the driver applies the brakes and brings the vehicle to rest over a distance of D m.
The lorry is then loaded with 800 kg of building materials that need to be transported over a total distance of 0.6 km. After loading, the lorry accelerates over a distance of 80m to attain a velocity of 30mph again. The vehicle subsequently maintains constant velocity before decelerating to rest at a rate of R m/s2.
Using only your own personal input data as given on the assignment 1 “input values” spreadsheet (penalties will result from not using your OWN data):-
a) Find for the unloaded lorry: ( 6 marks)
i) the force required to accelerate the vehicle, and the retarding force necessary to bring it to rest;
ii) the total time and the total distance traveled during this period.
b) Find for the loaded lorry : ( 7 marks)
i) the force required to accelerate the vehicle and the retarding force necessary to bring it
to rest;
ii) the total time taken during this period and the distance traveled at constant velocity.
c) Calculate the % change in the momentum of the lorry from loaded to unloaded, whilst traveling at the specified constant velocities. ( 3 marks)
d) Draw the ‘velocity-time’ graph of the motion of the lorry in its unloaded and loaded states, indicating clearly salient points and values. ( 4 marks) Relevant equations:
S=(u+v/2)t S = Displacement V = velocity A= acceleration
F= ma
Any help on this question would be extremely grateful as its just a little assignment i need to do - i don't have a clue how too :S I also have to sketch a VxT graph but i will be able to handle that haha :P any help would be grateful cheers guys
Regards mark
The Work i have done so far is
a = (v-u/t) = 30 - 0 /10 = 3ms^2
F = mxa = 2700 x 3 = 8100 N
8100 is the force required to accelerate?
Retarding force : m(-a) = 2700 x (-3) = -8100N ??
Total time.. = 50 seconds? distance traveld 15x50/2 = 375m ??
Am i correct or have i got it totally wrong?
Person_Number M (kg) D (m) R (m/s2)
<431383> 2700 60 1.4
Mechanical Systems Engineering
Assignment 1
An unloaded lorry of mass M kg accelerates from rest and reaches a velocity of 30mph in 10 seconds. It then travels at constant velocity for half a minute before the driver applies the brakes and brings the vehicle to rest over a distance of D m.
The lorry is then loaded with 800 kg of building materials that need to be transported over a total distance of 0.6 km. After loading, the lorry accelerates over a distance of 80m to attain a velocity of 30mph again. The vehicle subsequently maintains constant velocity before decelerating to rest at a rate of R m/s2.
Using only your own personal input data as given on the assignment 1 “input values” spreadsheet (penalties will result from not using your OWN data):-
a) Find for the unloaded lorry: ( 6 marks)
i) the force required to accelerate the vehicle, and the retarding force necessary to bring it to rest;
ii) the total time and the total distance traveled during this period.
b) Find for the loaded lorry : ( 7 marks)
i) the force required to accelerate the vehicle and the retarding force necessary to bring it
to rest;
ii) the total time taken during this period and the distance traveled at constant velocity.
c) Calculate the % change in the momentum of the lorry from loaded to unloaded, whilst traveling at the specified constant velocities. ( 3 marks)
d) Draw the ‘velocity-time’ graph of the motion of the lorry in its unloaded and loaded states, indicating clearly salient points and values. ( 4 marks) Relevant equations:
S=(u+v/2)t S = Displacement V = velocity A= acceleration
F= ma
Any help on this question would be extremely grateful as its just a little assignment i need to do - i don't have a clue how too :S I also have to sketch a VxT graph but i will be able to handle that haha :P any help would be grateful cheers guys
Regards mark
The Work i have done so far is
a = (v-u/t) = 30 - 0 /10 = 3ms^2
F = mxa = 2700 x 3 = 8100 N
8100 is the force required to accelerate?
Retarding force : m(-a) = 2700 x (-3) = -8100N ??
Total time.. = 50 seconds? distance traveld 15x50/2 = 375m ??
Am i correct or have i got it totally wrong?
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