(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

A 1.5kg block is at rest on a table. The coefficient of friction between the block and the table is 0.15. A force of 6N is applied to the block for 2 seconds followed by a force of 1N (in the same direction) for 3s.

(a)What is the speed of the block at the end (t=5)?

(b)What is the total distance the block moves up to t=5?

m=1.5kg

v_{1}=0m/s

CofF=0.15

F_{1}=6N

t_{1}=2s

F_{2}=1N

t_{2}=3s

2. Relevant equations

I really don't even have an idea which formulas to use

F=ma

F_{f}=(coefficient of friction)(F_{N})

E_{k}=(mv^{2})/2

W=FdcosA

v_{f}=v_{i}+at

d=v_{i}t+1/2at^{2}

2ad=v_{f}^{2}-v_{i}^{2}

d=vt

3. The attempt at a solution

F_{f}=CofF(F_{N})=0.15*1.5kg*9.8m/s^{2}=2.205N

F_{net}=ma --> a=F/m --> a=(F_{1}+F_{2}-F_{f})/m

a=3.1967m/s^{2}

??????????????

I only (really) need help with part (a). I've tried about a dozen solutions and none of them have worked. I've been working on this question for hours,PLEASEhelp.

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# Homework Help: Speed of block after 2 forces acted on it (Mechanics)

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