Question

A 2.0 kg mass starts from the rest and slides down an inclined plain 8.0*10^-1 m long on 0.50s. what net force is acting on the mass along the incline?

Answer 1

F=ma

Answer 2

we are given the mass and have to find a

Answer 3

but how do you find a.. do we just multiply by 9.8? or ?

Answer 4

we can find a using one of the four kinematics equations
\[v=v _{0}+at\]\[x=v _{o}t+\frac{ 1 }{ 2 }at ^{2}\] \[v ^{2}=v _{0}^{2}+2xa\]\[x=\frac{ 1 }{ 2 }(v+v_{0})t\]

Answer 5

we are given time and a length

Answer 6

so we use the equation that has time and length and leaves a as our only unknown

Answer 7

another important given is that it starts from rest so in all equations \[v_{0}=0\]

Answer 8

so, do use 2nd equation?

Answer 9

yes

Answer 10

i got 1.225 when i solved the equation?

Answer 11

our incline is .8m and it takes .5s\[.8=\frac{ 1 }{ 2 }a(.5)^{2}\]\[1.6=a(.5)^{2}\]\[1.6=.25a\]

Answer 12

a=6.4

Answer 13

opps, i didn't included .8

thanks

Answer 14

now you know a and m so just use

F=ma

and you will get your answer

Answer 15

12.8 N

Answer 16

technically 13 N