A student wishes to determine the coefficient of static friction between a block
ID: 1963050 • Letter: A
Question
A student wishes to determine the coefficient of static friction between a block of material and a wooden board. He places the board on a table, places the block on the board, and gradually raises one end of the board. When the end of the board has been raised 20 cm the block slides 77.3 cm down the entire length of the board in 1.6 seconds. Find: the coefficient of static friction; the coefficient of kinetic friction; the angle for which the velocity of the block will be constant. How hard must one press on the block, perpendicular to the inclined surface to prevent it from sliding down a 30 degree incline?Explanation / Answer
here ,
sintheta = Y/L
sin theta = 20/77.3
theta = 15 deg
when the frictional force is equal to gravitational force
so
us*m*g*costheta = m*g*sintheta
us = tan theta = tan15
us = 0.267 is the coefficient of static friction (ANSWER)
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(b) Length = distance S = 0.773 m , t = 1.6 s
so S = vo*t + 0.5* a*t^2
0.773 = 0 + 0.5*a*1.6^2
a = 0.603 m/s^2
Fnet = m*g*sin15 - uk*m*g*cos15 = m*a
9.8*sin15 - uk*9.8*cos15 = 0.603
uk = 0.204 is the coefficient of kinetic friction (ANSWER)
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(c) if velocity is constant , accceleration = a = 0
then
g*sintheta - uk*costheta = 0
tantheta = uk
theta = 11.5 deg is the angle for which the velocity of the block will be constant
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(d) N = (F+m*g*cos30)
m*g*sin30 = fs = us*N
m*g*sin30 = 0.246*(F+m*g*cos30)
F = (m*g*sin30 - us*m*g*cos30)/us
F = m*g*(sin30-0.267*cos30)/0.267
F = m*g so need to apply a force equal to the weight of the block
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