A wooden box weighing 10 kg is pulled evenly over a horizontal wooden board using

A wooden box weighing 10 kg is pulled evenly over a horizontal wooden board using a horizontal spring. The extension of the spring is 0.2 m. The coefficient of friction is 0.4. What is the stiffness of the spring?

m = 10 kg.

x = 0.2 m.

μ = 0.4.

g = 9.8 m / s2.

k -?

According to Newton’s first law, a body moves uniformly in a straight line if no forces act on it or the action of forces is compensated.

In the horizontal direction, the force F with which the bar is pulled is equal to the friction force Ff: Ff = F.

F = k * x.

In the vertical direction, the gravity force m * g is compensated with the surface reaction force N: m * g = N.

Ftr = μ * N = μ * m * g.

k * x = μ * m * g.

k = μ * m * g / x.

k = 0.4 * 10 kg * 9.8 m / s2 / 0.2 m = 196 N / m.

Answer: the spring has a stiffness k = 196 N / m.