Kinetic energy (Force and d(x))

AI Thread Summary
When no energy is dissipated, kinetic energy (KE) gained by an object can be calculated using the formula for work, which is the product of force and displacement. For a 10N force applied over 5m, the KE gained is 50 Joules, while a 20N force applied over 2m results in 40 Joules of KE. The discussion confirms that multiplying force (in Newtons) by displacement (in meters) yields the energy in Joules. This approach effectively demonstrates the relationship between work and kinetic energy. Understanding this principle is crucial for solving related physics problems.
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Homework Statement



If no energy is dissipated , how much KE does an object gain if
a) A 10N force is applied for 5m
b) A 20N force is applied for 2m

Homework Equations


The Attempt at a Solution


I know there is work being done if an object moves causing it to have KE. I think work is Force * displacement and the units for Force (N) and displacement (m) works out to Joules. So does that mean I use multiply force and displacement to find the amount of KE gained?
 
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Yes.
 
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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