What Determines the Thrust of a Lorry Starting from Rest?

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The effective thrust exerted by the lorry starting from rest with an acceleration of 2 m/s² can be calculated using the formula F=ma, where "m" is the mass of the lorry. Given a mass of 28,000 kg, the thrust is calculated as 28,000 kg multiplied by 2 m/s², resulting in 56,000 N. The discussion clarifies that the acceleration due to gravity does not affect the horizontal motion of the lorry on a level road surface. Therefore, only the horizontal acceleration is relevant for determining thrust. The final thrust exerted by the lorry is 56,000 N.
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Homework Statement


The lorry starts from rest with an acceleration of 2 ms-2.
What is the effective thrust is being exerted?

Homework Equations


acceleration=2 ms-2
starts from rest so initial velocity =0 ms-1

The Attempt at a Solution


F=ma
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You'll need to provide the mass of the lorry. That will be the "m" in your f = ma.
 
gneill said:
You'll need to provide the mass of the lorry. That will be the "m" in your f = ma.
The mass is 28000kg,so 28000x2=56000 N
would i just multiply it by the acceleration of 2 ms-2 but what about the acceleration due to gravity
 
Last edited:
fleur said:
The mass is 28000kg,so 28000x2=56000 N
would i just multiply it by the acceleration of 2 ms-2 but what about the acceleration due to gravity
Presumably the motion that is of interest here is in the horizontal direction, and the lorry is accelerating along a level road surface. Gravity operates in the vertical direction, and the road surface prevents motion in that direction. The vertical force due to gravity (weight) does not contribute to motion in the horizontal direction.
 
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Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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