With the help of a horizontal spring, a box with books is evenly pulled along the floor, overcoming a friction force of 5 N.

With the help of a horizontal spring, a box with books is evenly pulled along the floor, overcoming a friction force of 5 N. The length of the spring increases from 15 to 25 cm. What is the stiffness of this spring?

Ftr = 5 N.

l0 = 15 cm = 0.15 m.

l = 25 cm = 0.25 m.

k -?

Since the box with the books is pulled evenly in a straight line, then, according to 1 Newton’s law, the action of forces on it is compensated.

The friction force Ffr is compensated by the elastic force of the spring Ffr, the force of gravity m * g is compensated by the surface reaction force N: Ffr = Ffr, m * g = N.

We express the elastic force Fcont according to Hooke’s law: Fcont = k * x, where k is the stiffness of the spring, x is the change in the length of the spring.

x = l – l0.

Ftr = k * (l – l0).

k = Ffr / (l – l0).

k = 5 N / (0.25 m – 0.15 m) = 50 N / m.

Answer: the spring rate is k = 50 N / m.