Work and Kinetic Energy of a sled

AI Thread Summary
A sled being pulled by a constant horizontal force experiences four major forces: gravity, normal force, kinetic friction, and the applied force. The discussion clarifies that there is no tension force acting on the sled itself. Participants agree on the presence of gravity and kinetic friction, while emphasizing the horizontal force as the driving factor. A diagram illustrating these forces helps visualize their interactions. Understanding these forces is crucial for analyzing the sled's motion and energy dynamics.
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First, let us consider a sled of mass m being pulled by a constant, horizontal force of magnitude F along a rough, horizontal surface. The sled is speeding up.

how many major forces are acting on the sled? would it be 4 forces? gravity, normal, tension and kinetic?
 
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Yes,there are 4 forces acting on the sled.That 'kinetic' hopefully stands for 'kinetic friction force'...What tension...?There's the force that pulls the sled,F,there's no tension.

Daniel.
 
Gravity - yes. Kinetic (friction force) - yes. No tension.. well none that's really on the sled. Normal force - yes. And finally, the horizontal force. You get a nice diagram of 4 forces all on every side of the sled.
 
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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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