Calculating Final Speed of a 5kg Vase

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To calculate the final speed of a 5 kg vase lifted 10 m with a constant force of 60 N, the net force must account for gravity, leading to F_net = F - mg. The acceleration is calculated as a = F_net/m = (60 N - 49 N)/5 kg = 2.2 m/s². Using the kinematic equation v² = u² + 2as, where u is the initial speed (0), the final speed is determined as v = √(0 + 2 * 2.2 * 10) = 6.63 m/s. The initial calculation of 15.49 m/s was incorrect due to not considering the gravitational force. The correct final speed of the vase is 6.63 m/s.
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final speed ??

A 5 kg vase is lifted vertically from rest to a distance of 10 m with a constant upward applied force of 60.0 N. Calculate

iii) the final speed of the vase

F=ma
60=5(a)
a=12

v^2 = U^2+2as
v^2=0+2(12)(10)
V^2 = 240
v= 15.49m/s

is the step correct~??

thanks for reply~
 
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bboycs said:
F=ma
60=5(a)
a=12
What about mg?
 


is that have to include mg??
 


bboycs said:
is that have to include mg??
Yes. You have three forces acting on a vase: F, mg & ma.
 


method_man said:
Yes. You have three forces acting on a vase: F, mg & ma.

then can you show me the step to get the final speed??
 
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