A boy slide from an inclined plane of length 30m from the top, if the mass of a boy is 50 kg and there is a negligilble friction, find the speed of a boy when he reach at the bottom of the plane?:.
a = dv/dt OR a = v*dv/dx
Use the second equation. You know what 'a' is and you know what x0 and x1 are. You know that v0 = 0 m/s... Just solve for v1
Solution:
a*dx = v*dv
--now integrate
a*(x1-x0) = (v1^2-v0^2)/2
a= 9.8 m/s^2
x1= 30 m
x0= 0 m
v0= 0 m/s
9.8*30=v1^2/2
Solve for v1: approx 24 m/s
Reminder: the mass of the boy does not matter. Acceleration due to gravity is always the same no matter the mass of the object.
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Verified answer
a = dv/dt OR a = v*dv/dx
Use the second equation. You know what 'a' is and you know what x0 and x1 are. You know that v0 = 0 m/s... Just solve for v1
Solution:
a*dx = v*dv
--now integrate
a*(x1-x0) = (v1^2-v0^2)/2
a= 9.8 m/s^2
x1= 30 m
x0= 0 m
v0= 0 m/s
9.8*30=v1^2/2
Solve for v1: approx 24 m/s
Reminder: the mass of the boy does not matter. Acceleration due to gravity is always the same no matter the mass of the object.