Block can slide along a table where the coefficient of friction

[jbot2222]

A 198g wood block is firmly attached to a very light horizontal spring, as shown in the figure below.





The block can slide along a table where the coefficient of friction is 0.287. A force of 23.1 N compresses the spring 17.7 cm. If the spring is released from this position, how far beyond its equilibrium position will it stretch on its first swing

the way i approached this:

1/2kx^2 - Ukmgx = 1/2mv^2 = UkmgD
where D is the distance it travels after acquiring kinetic energy.

however, there is something wrong with this approach and i cannot figure out what it is. please help!

 

[Andrew Mason]

the way i approached this:

1/2kx^2 - Ukmgx = 1/2mv^2 = UkmgD
where D is the distance it travels after acquiring kinetic energy.

You are omitting the spring potential energy when the mass has extended to the distance D past the equilibrium position. This is provided by the original spring potential which also provides the work done against friction.

\frac{1}{2}k(x^2-D^2) =\mu_kmg(x+D)

AM

 

[wais]

Your approach is correct, but there are a few things to consider in this problem. Firstly, the equation 1/2kx^2 represents the potential energy stored in the spring when it is compressed by a distance x. This potential energy is then converted into kinetic energy as the spring is released and the block starts to move.

In this problem, the force of 23.1 N compresses the spring by 17.7 cm. This means that the block will have a potential energy of (1/2)k(0.177)^2 = 0.0157k when the spring is released.

Next, we need to consider the frictional force acting on the block as it slides along the table. The frictional force is given by Ff = ukmg where uk is the coefficient of friction, m is the mass of the block, and g is the acceleration due to gravity. In this case, Ff = (0.287)(0.198)(9.8) = 0.54 N.

Now, as the block starts to move, it will initially have a velocity of 0 since it was at rest. The kinetic energy acquired by the block as it moves a distance D is given by (1/2)mv^2. We can equate this to the potential energy stored in the spring and solve for D:

(1/2)mv^2 = 0.0157k - FfD

Substituting in the values we know, we get:

(1/2)(0.198)(v^2) = 0.0157k - (0.54)(D)

Solving for D, we get D = 0.0286 m or 2.86 cm. This is the distance the block will travel beyond its equilibrium position on its first swing.

Hope this helps!