# Results presentation mass of the wood block 1569 g

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Chapter 11 / Exercise 18
Physics for Scientists and Engineers
Jewett/Serway
Expert Verified
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Chapter 11 / Exercise 18
Physics for Scientists and Engineers
Jewett/Serway
Expert Verified
Mass(g) Newton(N) Fs(N) Fk(N) 500g+156.9g =656.9g6.44 N2.652 N1.895 N1000g+156.9g =1156.9g11.34 N5.385 N2.740 N 1500g+156.9g =1656.9g16.24 N 6.370 N 4.253 N 2000g+156.9g =2156.9g21.14 N 8.651 N 4.580 N 2500g+156.9g =2656.9g26.04 N 10.330 N 5.952 N 3000g+156.9g =3156.9g30.94 N 11.930 N 6.936 N Step 7:µs = 0.384µk = 0.218Step 8: CalculatedMeasured Differ %Fs(N)4.620 N4.026 N13.7%Fk(N)4.19 N 4.33 N 3.2%Step 10: Mass (g) µsƟ500g + block(m) 0.49261000g + block(m) 0.4223Step 11: Mass (g) µkƟ500g + block(m) 0.49261000g + block(m) 0.4022Final Result Type Friction coefficient Pulling MethodSliding Method%DifferStatic µs0.384 0.4924.1%Kinetic µk0.2180.4059.4%
Sample Calculations/ Graphs
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Discussion of Results:The main objective of this lab was to measure the coefficients of static and kinetic friction between a wooden block and a wooden plane. In this laboratory we investigated the kinetic and static friction forces acting on a wood block and on other materials. Results According to our hypothesis of Ff = Fn we were expecting a linear relationship between the force of friction and the normal force. This is what we obtained from both graphs.