Calculate the work done by the force

AI Thread Summary
To calculate the work done by the force on a particle moving along the curve defined by y=ax^2 - bx, where a=2.0 m^-1 and b=4.0 m, the force is given as F=cxy i + d j with c=8.0 N/m^2 and d=16 N. The work done can be expressed as the integral of the force over the path, W=∫F·ds. The next step involves calculating the integral by substituting the expressions for F and the differential path element ds. This requires evaluating the integral with respect to the defined limits from the origin to the point (3m, 6m). The discussion emphasizes the importance of setting up the integral correctly to find the total work done.
pph1011
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Homework Statement



A particle moves from the origin to the point x=3m, y=6m along the curve y=ax^2 - bx , where a = 2.0m^-1 and b = 4.0m . It is subject to a force F=cxy i + d j , where c= 8.0N/m^2 and d= 16N .

Homework Equations



Calculate the work done by the force

The Attempt at a Solution

 
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pph1011 said:

Homework Statement



A particle moves from the origin to the point x=3m, y=6m along the curve y=ax^2 - bx , where a = 2.0m^-1 and b = 4.0m . It is subject to a force F=cxy i + d j , where c= 8.0N/m^2 and d= 16N .

Homework Equations



Calculate the work done by the force

The Attempt at a Solution


Work is defined as:

\int\vec{F}d\vec{x}. Do you know what's next?
 
what's the next?
 
pph1011 said:
what's the next?

Take the integral of work.

W=\int Fds \rightarrow W = \int\int F dx dy

Catch my drift?
 
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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