Speed of block after 2 forces acted on it (Mechanics)

[Tasha9000]

Homework Statement

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
v₁=0m/s
CofF=0.15
F₁=6N
t₁=2s
F₂=1N
t₂=3s

Homework Equations

I really don't even have an idea which formulas to use
F=ma
F_f=(coefficient of friction)(F_N)
E_k=(mv²)/2
W=FdcosA
v_f=v_i+at
d=v_it+1/2at²
2ad=v_f²-v_i²
d=vt

The Attempt at a Solution

F_f=CofF(F_N)=0.15*1.5kg*9.8m/s²=2.205N

F_net=ma --> a=F/m --> a=(F₁+F₂-F_f)/m

a=3.1967m/s²

?

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, PLEASE help.

 

[N-Gin]

Be careful with the net force. You have to separate two different events: first you have force $$F_{1}=6 \textsc{N}$$ acting for 2 seconds and then $$F_{2}=1 \textsc{N}$$ for 3 seconds.

Now, it's not clearly stated if those forces are applied together after time interval of 2 seconds. If we assume that they are applied one after another then you have

$$F_{net_{1}}=F_{1}-F_{fr}=ma_{1}$$
$$F_{net_{2}}=F_{2}-F_{fr}=ma_{2}$$

From there we get accelerations. Distances and speeds are then easily calculated using given time intervals.

Otherwise, we have

$$F_{net_{1}}=F_{1}-F_{fr}=ma_{1}$$
$$F_{net_{2}}=F_{1}+F_{2}-F_{fr}=ma_{2}$$

 

[Tasha9000]

Thank you, you are a life saver