Question

In your physics lab, a block of mass m is at rest on a horizontal surface. You attach a light cord to the block and apply a horizontal force to the free end of the cord. You find that the block remains at rest until the tension T in the cord exceeds 20. 0 N. For T > 20. 0 N , you measure the acceleration of the block when T is maintained at a constant value, and you plot the results (Figure 1). The equation for the straight line that best fits your data is a=[0. 182m/(N⋅s2)]T−2. 842m/s2.

PART A: For this block and surface, what is the coefficient of static friction.


PART B: For this block and surface, what is the coefficient of kinetic friction

Answer 1

Coefficient of static friction: approximately
`0.371`.

Coefficient of kinetic friction: approximately
`0.290`.

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

Step-by-step explanation:

When the block is accelerating, the net force on this block is equal to
`T - (\text{kinetic friction})`. (
`T` is the tension the cord exerted on the block.)

Divide the net force on the block by the mass
`m` of the block to find the equation for acceleration
`a`:

`\begin{aligned} a &= \frac{(\text{net force})}{m} \\ &= \frac{T - (\text{kinetic friction})}{m}\end{aligned}`.

Rearrange this equation to obtain:

`\displaystyle a = \left((1)/(m)\right)\, T - \frac{(\text{kinetic friction})}{m}`.

In comparison, the equation of the best-fit line is:

`a = 0.182\, T - 2.842`.

The coefficient of
`T` (the slope) in the two equations need to match. Therefore:

`\begin{aligned}(1)/(m) &= 0.182\; ({\rm m\cdot N^(-1)\cdot s^(-2)}) \\ &= 0.182\; ({\rm m\, (kg \cdot m\cdot s^(-2))^(-1)\, s^(-2)}) \\ &= 0.182\; {\rm kg^(-1)}\end{aligned}`.

`\begin{aligned}m &= \frac{1}{0.182\; ({\rm kg^(-1)})} \approx (500/91)\; {\rm kg}\end{aligned}`.

Similarly, the vertical intercepts also need to match:

`\begin{aligned}\frac{(\text{kinetic friction})}{m} = 2.842\; ({\rm m\cdot s^(-2)})\end{aligned}`.

For part A, note that the maximum static friction on the block is the force that needs to be overcome for the block to start moving. In this question, the magnitude of this force is
`20.0\; {\rm N}`.

The magnitude of the normal force is the same as that of the weight of the block. The mass of the block is
`m \approx (500 / 91)\; {\rm kg}`. With
`g = 9.81\; {\rm N\cdot kg^(-1)}`, the weight of this block would be
`m\, g \approx 53.90\; {\rm N}`.

Divide the maximum static friction by the magnitude of the normal force to find the coefficient of static friction:

`\begin{aligned}\mu_{\text{s}} &= \frac{(\text{max static friction})}{(\text{normal force})} \\ &\approx \frac{20.0\; {\rm N}}{53.90\; {\rm N}} \\ &\approx 0.371\end{aligned}`.

Unlike static friction, as long as the block is moving on the same surface, the magnitude of kinetic friction would be constant.

For part B, divide kinetic friction by the normal force to find the coefficient of kinetic friction:

`\begin{aligned}\mu_{\text{s}} &= \frac{(\text{kinetic friction})}{(\text{normal force})} \\ &= \frac{(\text{kinetic friction})}{(\text{weight})}\\ &= \frac{(\text{kinetic friction})}{m\, g} \\ &= \left(\frac{(\text{kinetic friction})}{m}\right)\, \left((1)/(g)\right)\\ &\approx \frac{2.842\; {\rm m\cdot s^(-2)}}{9.81\; {\rm m\cdot s^(-2)}} \\ &\approx 0.290\end{aligned}`.