15. (ufpe) a figura a seguir mostra um bloco de massa 10 kg,
apoiado sobre uma superfície horizontal. ao longo da dire-
ção horizontal, indicada pelo eixo x, o bloco encontra-se sob
a ação de uma força constante de módulo fe de uma força
constante de módulo 30 n no sentido oposto. a equação
horária da posição do bloco é dada pela expressão x =
= 150 + 12t -0,60t2, onde x é dado em metros, et é dado
em segundos. qual é o valor de f em newtons?

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 and two forces acting on it. which are and in the opposite direction, so the net force acting on the object is:

(1)

In addition:

(2)

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

(3) Being the position in meters and the time in seconds.

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

(4)

(5) This is the velocity

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

(6)

(7) This is the acceleration

Substituting (7) in (2):

(8)

(9) This is the value of the net force

Substituting (9) in (1):

(10)

Finally finding :

(11)

b would be the answer

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explanation: