Question

When a 3.2 kg block is pushed against a massless spring of force constant 4.5 ✕ 103 N/m, The spring is compressed 7.9 cm. The block is released, and it slides 2.1 m (from the point at which it is released) across a horizontal surface before friction stops it. what is the coefficient of kinetic friction between the block and the surface?

Answer 1

To find the coefficient of kinetic friction between the block and the surface, we need to calculate the work done against friction. Using the given values and formulas for potential energy and work, we can determine the desired coefficient to be 0.35.

Step-by-step explanation:

To find the coefficient of kinetic friction between the block and the surface, we need to calculate the work done against friction. First, let's calculate the spring potential energy using the formula:

PE = 0.5 k x^2

where PE is the potential energy, k is the spring constant, and x is the displacement of the spring. Substituting the values, we get:

PE = 0.5 * (4.5 * 10^3) * (0.079)^2 = 22.45 J

Next, we can calculate the work done by friction using the formula:

W = F * d

where W is the work, F is the force of friction, and d is the distance. Substituting the values, we get:

W = μk * m * g * d

where μk is the coefficient of kinetic friction, m is the block's mass, g is the acceleration due to gravity, and d is the distance. Rearranging the formula, we can solve for μk:

μk = W / (m * g * d)

Substituting the known values, we get:

μk = W / (3.2 * 9.8 * 2.1)

μk = W / 64.128

Therefore, the coefficient of kinetic friction between the block and the surface is 22.45 / 64.128 = 0.35.

Answer 2

Approximately
`0.21` (assuming that
`g = 9.81\; {\rm N\cdot kg^(-1)}`.)

Step-by-step explanation:

During this motion, the elastic potential energy in the spring is converted into kinetic energy of the block, which is then lost because of friction. The amount of work that friction did on the block would be equal to the change in the kinetic energy of the block.

The coefficient of kinetic friction is the ratio between the magnitude of kinetic friction and the normal force between the two surfaces. Since the block is on a horizontal surface, the normal force between the surface and the block would be equal in magnitude to the weight of the block.

The coefficient of kinetic friction can be found in the following steps:

• Find the amount of elastic potential energy initially stored in the spring.
• Deduce the amount of work that friction did on the block. Divide the work by the displacement in the direction of the force to find the magnitude of friction on the block.
• Divide the magnitude of the kinetic friction on the block by the magnitude of normal force to find the coefficient of kinetic friction.

The amount of elastic potential energy (
`\text{EPE}`) stored in an ideal spring is:

`\displaystyle (\text{EPE}) = (1)/(2)\, k\, x^(2)`,

Where:

• 
  `k` is the spring constant, and
• 
  `x` is the displacement of the spring from equilibrium position.

In this question,
`k = 4.5 * 10^(3)\; {\rm N\cdot m^(-1)}` while
`x = 7.9\; {\rm cm} = 0.079\; {\rm m}` (note the unit conversion from centimeters to meters.)

Immediately after the block leaves the spring, the kinetic energy of the block would be equal to the maximum value of
`(\text{EPE})`, assuming that the effect of friction is negligible while the spring is decompressing.

`(\text{initial KE}) = (\text{EPE}) = (1/2)\, k\, x^(2)`.

The block stops after travelling
`2.1\; {\rm m}`, such that the final kinetic energy of the block would be
`(\text{final KE}) = 0`. The magnitude of the work that friction did on this block would be equal to the change in kinetic energy:

`\displaystyle W(\text{friction}) = (\text{final KE}) - (\text{initial KE}) = -(1)/(2)\, k\, x^(2)`.

Since friction opposes the motion of the block, the direction of friction on the block is opposite to that of the displacement. Hence, the distance travelled in the direction of friction would be negative:
`s = (-2.1)\; {\rm m}`.

Divide the work that friction did on the block by the distance travelled in the direction of friction to find the magnitude of friction on the block:

`\begin{aligned} (\text{kinetic friction}) &= \frac{W(\text{friction})}{s} \\ &= (-(1/2)\, k\, x^(2))/(s)\end{aligned}`.

Since the surface is horizontal, the normal force between the surface and the block would be equal in magnitude to the weight of the block,
`m\, g`.

Divide the magnitude of kinetic friction by the magnitude of the weight of the block to find the coefficient of kinetic friction:

`\begin{aligned}\mu_{\text{k}} &= \frac{(\text{magnitude of kinetic friction})}{(\text{magnitude of normal force})} \\ &= ((-1/2)\, k\, x^(2) / s)/(m\, g) \\ &= -(k\, x^(2))/(m\, g\, s) \\ &= -((4.5 * 10^(3))\, (0.079)^(2))/((3.2)\, (9.81)\, (-2.1)) \\ &\approx 0.21\end{aligned}`.

In other words, the coefficient of static friction between the block and the surface is approximately
`0.21`.