Calculating Work Done by Force on a Block

AI Thread Summary
The discussion revolves around calculating the work done by a force on a 100 kg block being pulled at a constant speed of 3 m/s. The applied force is 145 N at an angle of 37 degrees above the horizontal. The user initially calculated work using kinetic energy equations, yielding 450 W, but questioned the validity of this result. They then correctly applied the power formula, resulting in a value of approximately 347.41 W, confirming that power is indeed the rate at which work is done. The conversation concludes with the user feeling satisfied after clarifying their understanding of the concepts involved.
lauriecherie
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Homework Statement



A 100 kg block is pulled at a constant speed of 3 m/s across a horizontal floor by an applied foce of 145 N directed 37 degrees above the horizontal. What is the rate at which the force does work on the block? _____ W

Homework Equations


work = force (dot product) displacement
work = .5m * v^2 (final) - .5m *v^2 (initial)


The Attempt at a Solution



i used the second equation and plugged in my mass as 100kg and my final velocity as 3. so i got .5(100)(3^2) = 450 W. That seems like an awful lot to me ?
 
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Its the rate at which force does work.
What is the rate at which work is done known as?
 
Is it power?
 
This is something else I've tried.

Power = work / time = force (dot product) displacement = force * average velocity * cos(37 desgrees). I cam out with 347. 41 W. Am I doing this correctly?
 
Solved. Thanks for jogging my memory!
 
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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