Calculating Stopping Distance Using the Work-Energy Theorem

AI Thread Summary
To calculate the stopping distance of a car using the Work-Energy Theorem, the frictional force is identified as the force doing the work. The equation can be expressed as W = -F(friction)*x, where F(friction) equals u(k)mg. By setting the work done by friction equal to the change in kinetic energy, the equation becomes -u(k)mg*x = 0 - (1/2)mv^2. Solving for x provides the stopping distance in terms of the car's speed "v", gravitational acceleration "g", and the coefficient of kinetic friction "u(k)". This approach effectively combines the concepts of work, friction, and energy to determine stopping distance.
VinceStolen
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Homework Statement



A driver in a car is on a level road traveling at a speed of "v". He puts on the brakes and they lock and skid rather than roll. I have to use the Work-Energy Theorem to give an equation for the stopping distance of the car in terms of "v". the acceleration of gravity "g" and the coefficient of kinetic friction "u(k)" between the tires and the road.


Homework Equations



W = EK(f) - EK(i)

The Attempt at a Solution



I attempted to use various formulas I have that use friction and gravity but came up to no success. I am hoping someone else knows what they are doing.
 
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VinceStolen said:

Homework Statement



A driver in a car is on a level road traveling at a speed of "v". He puts on the brakes and they lock and skid rather than roll. I have to use the Work-Energy Theorem to give an equation for the stopping distance of the car in terms of "v". the acceleration of gravity "g" and the coefficient of kinetic friction "u(k)" between the tires and the road.


Homework Equations



W = EK(f) - EK(i)

The Attempt at a Solution



I attempted to use various formulas I have that use friction and gravity but came up to no success. I am hoping someone else knows what they are doing.
You have identified the work done in your formula. What force does this work? How do you calculate it? What is the definition of work?
 
The frictional force is the force doing this work. So W = -F(friction)*x. And F(friction) = u(k)mg. So -u(k)mg*x = 0 - (1/2)mv^2 ... and solve for x?
 
VinceStolen said:
The frictional force is the force doing this work. So W = -F(friction)*x. And F(friction) = u(k)mg. So -u(k)mg*x = 0 - (1/2)mv^2 ... and solve for x?
Looks good!
 
Thank you so much. You were extremely helpful.
 
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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