Finding Work and Power in a Traffic Junction

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A car weighing 2 tons approaches a traffic junction at 45 km/h and must accelerate to clear the intersection before the light turns red in 6 seconds. The driver needs to calculate the power output of the engine, factoring in a frictional resistance of 0.6 kN per ton. The initial velocity is converted to 12.5 m/s, while the final velocity required to cover 100 meters in the given time is 16.67 m/s. The discussion highlights the need for using kinematic equations to accurately determine acceleration and forces involved, as the initial calculations were incorrect due to not accounting for acceleration. The correct power output calculated is 82.9 kW.
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Homework Statement


A car of 2 ton mass traveling at 45 kmph approaches a traffic junction. When the car is 100m before the signal, the driver realizes that the green light is about to turn into red in 6 seconds and hence, he accelerates the car uniformly and crosses just before the light turns red. If the frictional resistance is 0.6 kN/ton, determine the power imparted by the engine at that instant.
Answer: 82.9 kW

Homework Equations


P= W/t ; W=Fd


The Attempt at a Solution


Vi = 45 kmph = 12.5 m/s
V= 100m/6sec =16.67 m/s
and i find the acceleration by the formula of V^2= Vi^2+2ax
and i use the formula of summation of forces horizontal : -Ff + F = ma
Ff= (0.6kN/ton)(2ton) = 1.2kN
and when i find F i substitute it to the formula of power F*100m , but my answer is wrong...please help...
 
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Your calculation for the velocity would be fine if the car was moving at a constant speed. Then, covering 100m in 6 seconds would yield an accurate value for the velocity. In this case, the car has an initial velocity and is accelerating. Do you have another kinematic equation in your repertoire that relates distance covered with initial velocity, acceleration, and time?
 
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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