Question

A 12 kg box sliding on a horizontal floor has an initial speed of 4.0 m/s. The coefficient of friction between the box and the floor is 0.20. The box moves a distance of 4.0 m in 2.0 s. The magnitude of the change in momentum of the box during this time is most nearly

Answer 1

Answer:

47.04 N*s or 47.04 kg*m/s

Explanation:

momentum is equal to mass times velocity the mass stays constant so we need to find the final and initial velocity. We are given the initial velocity of 4.0 m/s so plugging that into the momentum formula that is an initial momentum of 48 kg*m/s.

Now we can solve for  the final velocity by using the kinematic formulas. First we must find the net force which is the friction so friction is equal to coefficient * normal force with the normal force equaling the force of gravity. Therefore we get a frictional force  of 23.52 N

Now to find the acceleration we use newtons second a=f/m and get acceleration is equal to 1.96 m/s^2.

Now just find the Vfinal; Vfinal= Viinital +acceleration* time.

since the acceleration is negative we get 0.08m/s. Now just plug in to find final momentum of .96kgm/s. Then find the difference to be -47.04 kg*m/s.

Alternatively if you know that impulse is equal to change in momentum and equals force times time. We know the friction force is 23.52N in the negative direction applied for 2 seconds so we once again get  -47.04 N*s which is the same unit as kg*m/s

however since the question just asks for magnitude we take the absolute value and get 47.04 N*s

Answer 2

Answer:

47.04 Ns or 47.04 kgm/s

Explanation: