Asked by Chinasaramokwu
A body of 10kg and initially at rest is subjected to a force of 20N through a distance of 10m.calculate the change in kinetic energy of the body.
Answers
Answered by
Chinasaramokwu
M=10kg
H=5M
G=10m/s^2
P.E=mgh
P.E=10m×5M×10m/s
^2=500Watts
H=5M
G=10m/s^2
P.E=mgh
P.E=10m×5M×10m/s
^2=500Watts
Answered by
Chinasaramokwu
M=10kg
H=5M
G=10m/s^2
P.E=mgh
P.E=10m×5M×10m/s
^2=500Watts
Since p.e=k.e
So it is 500Watts
H=5M
G=10m/s^2
P.E=mgh
P.E=10m×5M×10m/s
^2=500Watts
Since p.e=k.e
So it is 500Watts
Answered by
Damon
Who said it was being lifted up?
work = force * distance = energy in
force = 20 N
distance = 10 m
work in = 20 N * 10 m = 200 Nm = 200 Joules = change in Ke if no change in height or energy sinks
Watts is Joules/second and unless you have typed this all wrong, power in watts has nothing to do with this.
work = force * distance = energy in
force = 20 N
distance = 10 m
work in = 20 N * 10 m = 200 Nm = 200 Joules = change in Ke if no change in height or energy sinks
Watts is Joules/second and unless you have typed this all wrong, power in watts has nothing to do with this.
Answered by
Damon
PHYSICS
Answered by
Chioma
Workdone = KE
F*d =1/2mv^2
20*10 =1/2*10*v^2
200=10v^2/2
400/10=V^2
v=√40
K.E=1/2mv^2
KE=1/2*10*√40
KE=200J
F*d =1/2mv^2
20*10 =1/2*10*v^2
200=10v^2/2
400/10=V^2
v=√40
K.E=1/2mv^2
KE=1/2*10*√40
KE=200J
Answered by
odey
great work
Answered by
Sofuwat
500
Answered by
Sofuwat
Nice job
Answered by
Halima muazu Ibrahim
Work =force×distance
Force=20n
Distance=10m
Work=20n×10m=200nm
Or 200joules
Change in kinetic energy =potential energy
The final answer =200joules
Force=20n
Distance=10m
Work=20n×10m=200nm
Or 200joules
Change in kinetic energy =potential energy
The final answer =200joules
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