Question

Brady is pulling a bag of candy around the house. If Brady's force remains constant and the bag rips and begins
losing candy what will happen to his acceleration?________________________________________________
Why?
If the bag is fixed and the candy in the bag doubles (Brady is applying the same force) what will
happen to the bags acceleration?

Answer 1

Use Newton's second law: F = m a, so a = F / m.

1) If Brady's force stays constant and the bag rips and loses candy (mass decreases) then the bag's acceleration increases.
Why? With the same F and smaller m, a = F/m gets larger (acceleration is inversely proportional to mass).

2) If the bag is fixed (same bag) and the candy in it doubles (mass doubles) while Brady applies the same force, the bag's acceleration is halved.
Reason: a = F/(2m) = (1/2)(F/m), so doubling the mass cuts acceleration by two.

(Real situations can show small deviations if the force or friction changes when the bag rips, but the F = ma relation gives the basic behavior.)