Work Done by the Gravitational Force

AI Thread Summary
The discussion revolves around calculating the work done by the gravitational force on a block of ice sliding down a frictionless ramp at a 50° angle. The user initially miscalculated the work done by applying the cosine function unnecessarily, leading to confusion about the energy difference. It was clarified that since the force exerted by the worker is opposite to the movement, the work done should simply be the product of the force and distance without needing to resolve components. The correct calculation indicates that the work done is 18 J, not 12 J, and the user acknowledges the misunderstanding. The conversation highlights the importance of understanding force direction and its impact on work calculations.
AtlBraves
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I can not figure out what I am doing wrong for this problem. Right now I have Wt = 60*.3*cos(50) = 12 J. If that is taken away, then it should be a 12 J difference right?

In Figure 7-33, a block of ice slides down a frictionless ramp at angle = 50°, while an ice worker pulls up the ramp (via a rope) with a force of magnitude Fr = 60 N. As the block slides through distance d = 0.30 m along the ramp, its kinetic energy increases by 80 J. How much greater would its kinetic energy have been if the rope had not been attached to the block?

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Work = Force * distance

Remember, you only need to multiply stuff by the cos (angle) or sin (angle) to get that component of force so it matches up with movement.

In this case, the 60 N applied in the opposite direction as the movement...so you don't need to find any of its components.

Simply, you would have a force of 60 N, and a distance of .30 m. With this info, you can easily find the energy.
 
AtlBraves said:
I can not figure out what I am doing wrong for this problem. Right now I have Wt = 60*.3*cos(50) = 12 J. If that is taken away, then it should be a 12 J difference right?
The force and the displacement are both parallel to the surface of the ramp.
 
I made that problem much harder than I should have. I understand now. Thanks for the help.
 
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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