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

1. Feb 10, 2010

### Tasha9000

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

2. Relevant equations

I really don't even have an idea which formulas to use
F=ma
Ff=(coefficient of friction)(FN)
Ek=(mv2)/2
W=FdcosA
vf=vi+at
d=vit+1/2at2
d=vt

3. The attempt at a solution

Ff=CofF(FN)=0.15*1.5kg*9.8m/s2=2.205N

Fnet=ma --> a=F/m --> a=(F1+F2-Ff)/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.

2. Feb 10, 2010

### 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

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

3. Feb 10, 2010

### Tasha9000

Thank you, you are a life saver