Work & Velocity: Solve 100m Distance, 2 Ton Car

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To determine the speed of a 2-ton car after traveling 100 meters with an initial speed of 80 km/h (22.22 m/s) and a frictional force of 1.2 kN, the deceleration due to friction must be calculated. The frictional force leads to a negative acceleration, which can be derived from the equation F = ma. Using the kinematic equation v^2 = u^2 + 2as, where 'u' is the initial speed, 'a' is the deceleration, and 's' is the distance, the final speed can be computed. The discussion emphasizes the importance of using the correct equations of motion rather than focusing on work calculations. Ultimately, applying these principles will yield the final speed of the car after 100 meters.
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Homework Statement


The driver of a car wishing to save the fuel takes his leg off the accelerator when the car is moving at a maximum speed of 80 kmph. Determine the speed of the car after traveling a distance of 100m. The mass of car is 2 tons and the force of friction is 0.6 kN/ ton.


Homework Equations


Work = Force * distance
kinetic force = 1/2 (mV^2 - mVi^2)



The Attempt at a Solution


Frictional force = 0.6kN/ton( 2 tons) = 1.2 kN
Vi= 80 kmph = 22. 22 m/s
i don't know how to start please help me ...please..
 
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use equation of motion
v^2 = u^2 + 2as
 
Why are you trying to calculate work? The problem does not ask for that. You have the friction force so can calculate the deceleration (negative acceleration). Use the formula that ashishsinghal gives.
 
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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