- #1
Benjamin_harsh
- 211
- 5
- Homework Statement
- A horizontal force of 150N is applied on a 20kg box which causes it to move to the right. What is the acceleration if the coefficient of kinetic friction is 0.25?
- Relevant Equations
- ##\sum F_{X} = F - F_k##, ##ma = F - \mu_{k}.F_k##
##\sum F_{X} = F - F_k##
(The net force, ##\sum F_{X}## is always equal to m.a)
##ma = F - \mu_{k}.F_n##
##ma = F - \mu_{k}.(mg)## [Here ##F_n = mg## when body is on flat surface]
##20(a) = 150 - 0.25(20)(10)##
##\large a = \frac {150 - 0.25(20)(10)}{20}##
##\large a = 5\frac{m}{sec^2}##
How ##\sum F_{X} = m.a## ?
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