Need to know how to solve for test

[kiarrahannice]

A 2.0kg box rests on a plank that is inclined at a angle of 65 degrees above the horizontal. The upper end of the box is attached to a spring with a force constant of 295 N/m. If the coefficient of the static friction between the box and the plank is 0.19, what is the maximum amount the spring can be stretched and the box remain at rest?

Can someone show me how to complete this problem. The answer is 6.6cm however I don't know to get there. thanks

I also know that I have to find force and divide that by 295N/M to get answer can someone help me find force please! What numbers do I use

 

[Delphi51]

Get expressions for the forces on the box in the direction of the plank. You'll need to resolve the mg force into a component along the plank and another perpendicular to the plank that you can use to find the friction forces. Set the sum of the forces equal to zero - the point where it will just begin to move. That should allow you to solve for the stretch.

 

[nlightner]

To solve this problem, we will use the principles of static equilibrium and Hooke's Law.

First, we need to draw a free body diagram of the box to identify all the forces acting on it. The weight of the box (mg) acts downwards, while the normal force (N) from the plank acts upwards. Additionally, there is a static friction force (fs) acting on the box in the opposite direction of its motion. The spring force (Fs) also acts upwards, as it is attached to the upper end of the box.

Next, we need to set up an equation for the sum of forces in the vertical direction. This will allow us to solve for the normal force.

ΣFy = N - mg = 0

N = mg

Now, we can set up an equation for the sum of forces in the horizontal direction. This will allow us to solve for the static friction force.

ΣFx = fs - Fs = 0

fs = Fs

To find the maximum amount the spring can be stretched, we need to find the maximum value of fs. This occurs when the box is on the verge of slipping, so the static friction force is at its maximum value. We can calculate this using the formula for static friction:

fs = μN

where μ is the coefficient of static friction. Substituting in our values, we get:

fs = 0.19(mg)

Next, we need to find the spring force (Fs) using Hooke's Law, which states that the force exerted by a spring is proportional to the distance it is stretched or compressed.

Fs = kx

where k is the force constant of the spring and x is the distance the spring is stretched. In this case, x is also the maximum amount the spring can be stretched before the box starts to move.

Substituting in our values, we get:

Fs = (295 N/m)(x)

Now, we can set up an equation for the sum of forces in the vertical direction to find the maximum amount the spring can be stretched:

ΣFy = fs + Fs - mg = 0

Substituting in our values, we get:

0.19(mg) + (295 N/m)(x) - mg = 0

Solving for x, we get:

x = (0.19mg)/(295 N/m)

Substituting in the value for mg (mass x gravity), we