Solving Energy Problems: Q on Acceleration, Distance, Force & Friction

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To solve the energy problem, the key points include calculating the distance moved by the car, which is confirmed to be 80 m, and determining its acceleration at 1.6 m/s². The force exerted by the motor can be calculated using the work-energy principle, where the work done (400 J) is divided by the distance (80 m), yielding a force of 5 N. The frictional force opposing the car's motion can be found by subtracting the work done to accelerate the car from the total work, indicating that friction does less than 400 J of work. Finally, the final kinetic energy of the car can be derived from the work done to accelerate it.
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My teacher assigned a few problems, and I'm having difficulty with one of them. The question is:

In order to accelerate a 2 kg car from rest to a speed of 16 m/s, 400 J of work are done by the motor of the car. The acceleration takes 10s. Calculate:

a) the distance moved by the the car (I got 80 m)
b) its acceleration (1.6 m/s2)
c) the force provided by the motor
d) the frictional force opposing the car's motion
e) the work done to overcome friction
f) the final kinetic energy of the car

I need help especially with c and d. Thanks.
 
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c) W=fd
f=W/d
 
d) find the work that was done on the car to get to 16m/s^2, it should be less than 400J, the difference is the work done by friction. Since the force is constant, just divide that work by the distance to get the force.
 
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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