What is the final speed of the box at 10m?

AI Thread Summary
A 2.0kg mass is lifted 10m with a force that varies from 20N to 30N, leading to a non-constant acceleration. To find the final speed at 10m, the work done by the force can be calculated as the area under the force vs. distance graph. This work is then equated to the gravitational potential energy and kinetic energy using the conservation of energy principle. The final speed of the box is determined to be 7.3m/s. Understanding the relationship between work, force, and distance is crucial for solving this problem.
avsj
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Homework Statement



A force is exerted by a rope lifting a 2.0kg mass a vertical distance of 10m from the ground. A graphs shows F (N) vs d (m). THe F starts at 20 N and goes to 30 N in a straight line as d goes from 0 to 10m. What is the final speed of the box at 10m?

Homework Equations



F=ma
kinematics equations

The Attempt at a Solution



I am thoroughly lost as to how to deal with this one. The force applied is not constant, so the acceleration will also not be constant. I'm guessing there is another way of looking at it... would really appreciate a hint in which direction to start... thanks.. the answer is 7.3m/s by the way
 
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Work is equal to the integral over distance of force. Use conservation of energy to compare this with mgh and then use KE=mv^2/2.
 
what do you mean work is equal to the integral over distance of force, what is the integral?
 
avsj said:
what do you mean work is equal to the integral over distance of force, what is the integral?

No calculus in this course? Then if you graph force versus distance, work is the area under the curve. That's what an integral is.
 
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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