# Under the action of a horizontal force equal to 12N, the body moves according to the law x = x0 + 1.01t2.

**Under the action of a horizontal force equal to 12N, the body moves according to the law x = x0 + 1.01t2. Find the mass of the body if the coefficient of friction is 0.1**

F = 12 N.

g = 10 m / s ^ 2.

μ = 0.1.

x = x0 + 1.01 * t ^ 2.

m -?

Let’s write 2 Newton’s law in vector form: m * a = F + Ffr + m * g + N.

ОХ: m * a = F – Ftr.

OU: 0 = m * g – N.

m * g = N.

The friction force Ffr is determined by the formula: Ffr = μ * N.

m * a = F – μ * m * g.

m * a + μ * m * g = F.

m * (a + μ * g) = F.

m = F / (a + μ * g).

The acceleration of the body a (t) will be the second derivative of the dependence of the coordinate on time: a (t) = x (t) “”.

a (t) = (x0 + 1.01 * t ^ 2) “” = (1.01 * 2 * t) “= 2.02 m / s ^ 2.

m = 12 N / (2.02 m / s ^ 2 + 0.1 * 10 m / s ^ 2) = 3.97 kg.

Answer: body weight is m = 3.97 kg.