A truck starting from rest moves down a hill with a constant acceleration .if it covers a distance of 300m in 10s .find its acceleration also find the force acting on it if its mass is 8000kg
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the truck starts from rest so u=0
s=ut+1/2 at'2
s=400, t =20, we should find acceleration
s=ut+1/2 at''2
400=[0]20+1/2[a] [20] [20]
400=0+1/2[400] [a]
400=[200] [a]
400/200=a
2=a
f=ma
[1tonne]=1000kg
then 7 tonne=[7] [1000]
m =7000
a=2
f=ma
f=[7000] [2]
=14000 N
hope it helped ya...:) :)
the truck starts from rest so u=0
s=ut+1/2 at'2
s=400, t =20, we should find acceleration
s=ut+1/2 at''2
400=[0]20+1/2[a] [20] [20]
400=0+1/2[400] [a]
400=[200] [a]
400/200=a
2=a
f=ma
[1tonne]=1000kg
then 7 tonne=[7] [1000]
m =7000
a=2
f=ma
f=[7000] [2]
=14000 N
hope it helped ya...:) :)
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