Why is work done by friction negative in this problem?

[xzibition8612]

Homework Statement

See attachment

Homework Equations

The Attempt at a Solution

Why is it -umg(50)? Why negative? Also I think this has something to do with conservative force and friction, and I'm very confused on this. Any help would be appreciated thanks.

 

[Curious3141]

Homework Statement

See attachment

Homework Equations

The Attempt at a Solution

Why is it -umg(50)? Why negative? Also I think this has something to do with conservative force and friction, and I'm very confused on this. Any help would be appreciated thanks.

When the force is constant, Work = Force*Displacement.

In this case, it's negative because energy is being taken away from the cars by the frictional force, while they're being decelerated to a stop. The energy dissipated by friction is equal to the drop in kinetic energy (as each car comes to rest). This is conservation of energy.

μ is the coefficient of kinetic friction. mg is the weight (equal to the normal force on the car). 50ft is the distance through which car 1 moves.

BTW, there's an error in the second part of the solution. The expression for car 2 should be $\sqrt{(2)(20)(0.5)(32.2)}$ (i.e. 20, not 50), since 20ft is the distance car 2 takes to be brought to rest. The final stopping distance was calculated correctly, though.