- #1
Andrew1234
- 18
- 1
- Homework Statement
- The problem is attached.
- Relevant Equations
- 1/2mv^2 = (Fb+Fk)d
The truck stops in a shorter distance if the crate slides but why is this the case considering that friction on the crate does positive work on the truck, since it points left for the crate but right for the truck thereby opposing the braking force?
I understand why using the equation
1/2mv^2 = (Fb+Fk)d, where Fb is the braking force and Fk the spring force, d is shorter if sliding occurs but not why Fk is directed opposite the motion given that it points opposite the crate's velocity, left, but must be balanced on the truck's free body diagram by an equivalent force that points to the right.
I understand why using the equation
1/2mv^2 = (Fb+Fk)d, where Fb is the braking force and Fk the spring force, d is shorter if sliding occurs but not why Fk is directed opposite the motion given that it points opposite the crate's velocity, left, but must be balanced on the truck's free body diagram by an equivalent force that points to the right.