- #1

Benjamin_harsh

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- Homework Statement
- A force of 100N applied to a body which has 5kg mass. Coefficient of static friction(##μ_{s}##) is 0.60 and Coefficient of kinetic friction(##μ_{k}##) is 0.55.

- Relevant Equations
- Acceleration from Net force, ##a = \large \frac{70.6}{5}## = ##14.1 \;m/sec^{2}## to the right.

Acceleration from Net force, ##a = \large \frac{73.06}{5}## = ##14.61 \;m/sec^{2}## to the right.

A force of 100N applied to a body which has 5kg mass. Coefficient of static friction(μs) is 0.60 and Coefficient of kinetic friction(μk) is 0.55.

A) ##μ_{s} = 0.60##; ##μ_{k} = 0.55##

##F_{s}^{max} = F_{n}*μ_{s}##

##F_{s}^{max} = 49*0.60 (F_{n} = 9.8 \;m/sec^{2} *(5kg) = 49N)##

##F_{s}^{max} = 29.4 N##

Net force, ##F_{net} = 100 – 29.4N = 70.6N##

Acceleration from Net force, ##a = \large \frac{70.6}{5}## = ##14.1 \; m/sec^{2}## to the right.

Frictional force = ##F_{n}*μ_{k}##

Frictional force = ##49*(0.55) = 26.95N##

Net Force, ##F_{net} = 100 – 26.95 = 73.05N##

Acceleration from Net force, ##a = \large \frac{73.06}{5}## = ##14.61 \; m/sec^{2}## to the right.

A) ##μ_{s} = 0.60##; ##μ_{k} = 0.55##

##F_{s}^{max} = F_{n}*μ_{s}##

##F_{s}^{max} = 49*0.60 (F_{n} = 9.8 \;m/sec^{2} *(5kg) = 49N)##

##F_{s}^{max} = 29.4 N##

Net force, ##F_{net} = 100 – 29.4N = 70.6N##

Acceleration from Net force, ##a = \large \frac{70.6}{5}## = ##14.1 \; m/sec^{2}## to the right.

Frictional force = ##F_{n}*μ_{k}##

Frictional force = ##49*(0.55) = 26.95N##

Net Force, ##F_{net} = 100 – 26.95 = 73.05N##

Acceleration from Net force, ##a = \large \frac{73.06}{5}## = ##14.61 \; m/sec^{2}## to the right.