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- Problem Statement
- A child pushes a block of wood with a mass of 0.72kg across a smooth table. The block starts from a position of rest and after 2.0 s it has a velocity of 1.6m/s [forward]. The coefficient of kinetic friction is 0.64

- Relevant Equations
- FN= Fg=mg , Fnet=Fg+FN, Ff=Fnuk

I am very new to physics so I am still learning a lot. Here is my attempt:

Find the net force acting on the block : Fnet= Fg+FN so I have to find FN before I can complete the answer. FN = Fg (mg) FN = (0.72kg)(9.8m/s^2) = 7.056 or 7.06 N

Fnet = 9.8+7.06 = 16.86 N or 16.9 N

To find the force of friction I used the formula Ff= (FN)(uk) = 16.9*.64 =10.816N or 10.8 N

Now I used the formula Fapplied = Ff+FN = 10.8N + 7.06N = 17.86 or 17.9 N of applied force.

Am I even close to understanding this or correct?

Find the net force acting on the block : Fnet= Fg+FN so I have to find FN before I can complete the answer. FN = Fg (mg) FN = (0.72kg)(9.8m/s^2) = 7.056 or 7.06 N

Fnet = 9.8+7.06 = 16.86 N or 16.9 N

To find the force of friction I used the formula Ff= (FN)(uk) = 16.9*.64 =10.816N or 10.8 N

Now I used the formula Fapplied = Ff+FN = 10.8N + 7.06N = 17.86 or 17.9 N of applied force.

Am I even close to understanding this or correct?