Reference no: EM13808905
1. Using the mass of the block and the average force of kinetic friction from Data Table 1, calculate the coefficient of kinetic friction from Equation 1:
Ffr(k) = uk FN. Therefore mk = Ffr(k) / FN
.87N= uk (4.11N)
uk= .87N/4.11N
uk= .211N
2. Using the mass of the block and the average force of kinetic friction from Data Table 2, calculate the coefficient of kinetic friction for the wood block sliding on its side. Record your results and see how it compares with the value of Uk obtained from Data Table 1.
Uk = Ffr(k) / FN
Uk= .88N/ 4.11N
Uk= .214N
It is very similar, which leads me to conclude that sliding friction is proportional to the mass of an object, not the surface area.
3. From the data in Data Table 3, 4, and 5, compute the coefficient of static friction, Us, for the glass surface on wood, the sandpapered surface on wood, and wood on carpet, etc., from each of your three trials. Calculate an average value of Us. Record your results in your own data sheets.
Glass
Us = Ffr(s) / FN
Us= .6N/4.11, 6.N/4.11N, .5N/4.11N
Us= .15N, .15N, .12N
Us average= .14
Sandpaper
Us = Ffr(s) / FN
Us= 1.7N/4.11N, 1.8N/4.11N, 1.7N/4.11N
Us= .41N, .44N, .41N
Us average= .42N
Carpet
Us = Ffr(s) / FN
Us= 2.4N/4.11N, 2.4N/4.11N, 2.3N/4.11N
Us= .58N, .58N, .56N
Us average= .57N
4. From the data obtained in Data Table 6 calculate Us for wood on wood from each of your three trials.
Us = tan θ = sin θ max/ cos θ max or Us = tan θ = height/base
Trial 1= tan 22.3= .41 or 25/61 = .41
Trial 2=tan 22.1= .41 or 26/64 = .41
Trial 3= tan 22= .40 or 25.5/63= .40
5. Calculate an average value of Us. Record your result on the data sheet.
(.41+.41+.40)/3 = .41
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