A box of mass M is pushed a distance Δ x across a level floor by a constant applied force F . The coefficient of kinetic friction between the box and the floor is μ . Assuming the box starts from rest, express the final velocity v f of the box in terms of M , Δ x , F , μ , and g .

Final speed of the box after moving the given distance is

Explanation:

By work energy theorem we know that work done by all the forces is equal to the change in kinetic energy of the system

Here we know that work is done on the system by external force and friction force

So we will have

so we have

graphing motion: mastery test

209be

what is the instantaneous acceleration of

the particle at point b?

0 meters/second?

10 meters/second?

12 meters/second?

velocity imetersisecond

15 meters/second?

20

time (seconds)

answer: 0 meters/second^2

force is mass*acceleration.

we have a very small mass: 11.4*0.001 kg

if we want 0.035n of force, we need an acceleration:

0.035 = 11.4*0.001*a

a = 0.035/(11.4*0.001) m/s^2