Power Calculation for 1100 kg Object at 5 m/s

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To calculate the power generated by pulling a 1100 kg object at a constant velocity of 5 m/s with a force of 122 N at a 37-degree angle, the basic formula for power, P = F x v, suggests a value of 610 watts. However, the discussion emphasizes the need for a deeper understanding of the forces involved, including the importance of free body diagrams to analyze the situation accurately. Participants highlight that calculating work using W = Fdcos(37) may be necessary, especially when considering the object's initial state. The conversation stresses the importance of showing one's own effort before receiving help with homework problems. Overall, a comprehensive approach is needed to solve the power calculation accurately.
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A 1100 kg object is pulled across a horizontal surface at a constant velocity of 5 m/s. What is the power generated by this force? (Picture includes object being pulled at a 37 degree angle above the horizontal with a Force of 122 N)
 
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Bangdeek27, welcome to PF!. Please read the forum rules and follow the format listed when you post. Homework helpers will not assist with any questions until you've shown your own effort on the problem. Remember, we help with homework, we don't do your homework. Please list what you believe are the relevant equations and show some attempt at a solution. Thanks.
 
Well I thought that power just = force x velocity, so (122)(5) = 610 watts. However I really don't think its that simple. I don't understand how to do any free body diagrams for this problem, though I feel they are needed.
 
bandgeek27 said:
Well I thought that power just = force x velocity, so (122)(5) = 610 watts. However I really don't think its that simple. I don't understand how to do any free body diagrams for this problem, though I feel they are needed.


Uhe formula for work: W = Fdcos37*, solve for d by assuming that the mass starts from rest.
 
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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