Work, Energy & Power: Solving 20 & 21 Problems

AI Thread Summary
To solve problem 20, the work done by a 22kW car engine in 60 seconds can be calculated using the formula Work = Power x Time, resulting in 1320 kJ if the engine is 100% efficient. For problem 21, part a requires calculating work done using the formula W = F x d, where the distance can be found by multiplying speed and time, leading to 150 J. Part b involves finding power, which is calculated as Power = Work / Time, yielding 6 W. Part c asks for the force of friction, which can be determined by subtracting the applied force from the net force, resulting in a frictional force of 5 N. The discussion emphasizes the application of fundamental physics equations to solve work, energy, and power problems effectively.
McKeavey
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Homework Statement


20. How much work can a 22kW car engine do in 60s if it is 100% efficient?

21. A force of 5.0N moves a 6.0 kg object along a rough floor at a constant speed of 2.5m/s.
a) How much work is done in 25s?
b) What power is being used?
c) What force of friction is acting on the object?



Homework Equations





The Attempt at a Solution


I would guess that I need to use W = F * d , but got nowhere.

W = F * d
P = E/t
got nowhere too..
 
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Work= Power x time. so much work would the car do, since P =22 kW and t=60 s ?

since its 100 % efficient, all the energy given by the engine is used for moving the car
 
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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