Question

15. (UFPE) A figura a seguir mostra um bloco de massa 10 kg,

Answer 1

Answer: 18 N

Explanation:

According to the described situation, the net force in the Y component is zero and we will work only with the X component.

We have a block with mass $m=10 kg$ and two forces acting on it. which are $F_{e}$ and $30N$ in the opposite direction, so the net force $F$ acting on the object is:

$F=F_{e}-30N$ (1)

In addition:

$F=m.a$ (2)

On the other hand, acceleration $a$ is the result of the second derivative of the position with respect to time, this means we have to derive twice the following equation:

$X=150 + 12 t - 0.60 t^{2}$ (3) Being $X$ the position in meters and $t$ the time in seconds.

Let's derive (3) and find the velocity $V$:

$V=\frac{d}{dt}(X)=\frac{d}{dt}(150) + \frac{d}{dt} (12 t) - \frac{d}{dt} (0.60 t^{2})$ (4)

$V=\frac{d}{dt}(X)=12 - 1.2 t$ (5) This is the velocity

Now, let's derive (5) and find the acceleration:

$a=\frac{d}{dt}(V)=\frac{d}{dt}(12) - \frac{d}{dt}(1.2 t)$ (6)

$a=-1.2m/s^{2}$ (7) This is the acceleration

Substituting (7) in (2):

$F=(10kg)(-1.2m/s^{2})$ (8)

$F=-12 N$ (9) This is the value of the net force

Substituting (9) in (1):

$-12 N=F_{e}-30N$ (10)

Finally finding $F_{e}$:

$F_{e}=18 N$ (11)