Question

i need to find the slope and whatever else they are asking

Calculating the Coefficient of Static Friction (Pinewood on painted steal) From Empirical Data Background: **An object at rest remains at rest unless a force is applied to it." This paraphrases a portion of Newton's First law of motion. In the case of this lab a minimum amount of force must be applied to overcome the attraction between two surfaces. The amount of force needed to start the motion is proportional to the normal force (equal in magnitude to a component al weight) and a constant (?) called the coefficient of friction His). If the coeficient is a constant than a plot of the force needed to move an object as a function of its weight should be a straight line, and its would be the slope. In the case of the first measurements, you will have a weighted box on a horizontal surface. A string hanging over a pulley to a weight hanger provides a source of tension to the box. The tension will equal the force due to the hanging weight. The force of friction will be the normal force (N) times : W = (us)(N) The second set of trials will be similar except the box will be on an incline plane. The normal force will be a component of the total weight. (us)(N) (M + mig No incline mag

Answer 1

Part A: The curve is plotted beween (m2g and N) to find the slope i.e. $\mu s$

Slope is given by value M in the graph below : M = $\mu s$ = 63.083 gm/N

M*X+B PART A: m2g v/s Force (N) Family Function Linear Parameter Estimates trace 0 Fit of trace 0 M 63.0833... B -100.26... Run Mass (grams) 2.6 3 3.2 3.4 3.4 3.6 3.6 3.8 3.8 4 Activate Vlandows Go to Settings to activate Windows. Force (Newtons)

Part B : Doing similar operation as done in part A:

Slope for the Table in Part B is equal to : 53.5833

M*X+B PART B: m2g v/s Force (N) Family Function Linear Parameter Estimates trace 0 Fit of trace 0 M 58.5833... 32.9277... Run Mass (grams) 2.6 2.6 2.8 3.2 3.4 3.4 3.6 3.6 3.8 3.8 4 Activate Windows Go to Settings to activate Windows. Force (Newtons)

Analysis: From the graph, it is clearly visible that the data points follow the straight line curve and the slope values are also numerically near by to each others.

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