A 50kg safe made of steel rests on a steel floor. A spring with a spring constant of 550 N/m is attached to the side of the safe and then stretched out to the right. If the spring starts out 1m long, how long will it stretch to before the safe moves? Draw a free body diagram and solve.

It will stretch 66 [cm]

Explanation:

In order to solve this problem, we must understand that there is movement between the safe and the steel floor, that is, there is a force of friction between surfaces. Since the value of static coefficient of friction between steel surfaces is not given, we must perform an Internet search in order to find an approximate value of this value.

μ = 0.74

In the attached image we can find the free body diagram with the forces interacting on the safe.

First we perform a summation of forces on the Y axis equal to zero, in this way we can find the value of the normal force, which will be needed later.

Then we perform a summation of forces on the X axis equal to zero, in order to find the equation that relates the frictional force with the elastic force of the spring.

Knowing that the friction force is equal to the product of the normal force by the coefficient of friction we can find the elastic force.

The elastic force in a spring is equal to:

Fe = k*x

Clearing x

x = 362.97 / 550

x = 0.66 [m] or 66 [cm]

Note: In the attached image we can see the free body diagram with the forces acting over the safe.

the force required to move the cart is divided by three or 33.33 n.

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explanation: