chris097
- 16
- 0
Homework Statement
The coefficient of static friction is 0.636 between the two blocks shown. The coefficient of kinetic friction between the lower block and the floor is 0.138. Force F causes both blocks to cross a distance of 7.00 m, starting from rest. What is the least amount of time in which the motion can be completed without the top block sliding on the lower block, if the mass of the lower block is 1.14 kg and the mass of the upper block is 2.80 kg?
http://capa.physics.mcmaster.ca/figures/kn/Graph08/kn-pic0832.png
The attempt at a solution
Fmax= UsN
=(.636)(2.80*9.81)
=17.4696 N
F(bottom) = UkN
= (.138)((2.80 + 1.14) * 9.81)
=5.3339
Fnet = 17.4696 - 5.3339
=12.1357
F=ma
a=F/m
=12.1357/(1.14+2.8)
=3.0801
d=v1*t + .5at^2
t=square root of (d/(.5a))
= square root of (7/(.5(3.0801)))
=2.13 s
where am i going wrong?
thank you
The coefficient of static friction is 0.636 between the two blocks shown. The coefficient of kinetic friction between the lower block and the floor is 0.138. Force F causes both blocks to cross a distance of 7.00 m, starting from rest. What is the least amount of time in which the motion can be completed without the top block sliding on the lower block, if the mass of the lower block is 1.14 kg and the mass of the upper block is 2.80 kg?
http://capa.physics.mcmaster.ca/figures/kn/Graph08/kn-pic0832.png
The attempt at a solution
Fmax= UsN
=(.636)(2.80*9.81)
=17.4696 N
F(bottom) = UkN
= (.138)((2.80 + 1.14) * 9.81)
=5.3339
Fnet = 17.4696 - 5.3339
=12.1357
F=ma
a=F/m
=12.1357/(1.14+2.8)
=3.0801
d=v1*t + .5at^2
t=square root of (d/(.5a))
= square root of (7/(.5(3.0801)))
=2.13 s
where am i going wrong?
thank you
Last edited by a moderator: