mg = kx so
k = mg/.05
Work done is 1/2kx^2 using k from before.
PS. I don't get 2J, I get 1J
An ideal spring hangs vertically from the ceiling. When a 1.0 kg mass hangs from the spring it is extended 5.0 cm from its relaxed length. A downward external force is now applied to the mass to extend the spring an additional 10 cm. While the spring is being extended by the force, the work done by the spring is:
The answer is : -2.0 J (But I keep getting -0.7 J) ..
Please explain - Thank you
2 answers
The guy up there is half right. But you have to take into account the force of gravity as well since the work done by the spring is going to be the energy from the external force and the energy from the force of gravity. So,
mg= kx
k=mg/0.05
work done is 1/2kx^2 which gives you 0.98J.
You then take into account grav. which is
F=mg
F=1(9.8)=9.8N
W=Fd
W=9.8(.1)=.98
.98+.98=1.96 which is 2, so the work done by the spring is just opposite that.
mg= kx
k=mg/0.05
work done is 1/2kx^2 which gives you 0.98J.
You then take into account grav. which is
F=mg
F=1(9.8)=9.8N
W=Fd
W=9.8(.1)=.98
.98+.98=1.96 which is 2, so the work done by the spring is just opposite that.
1 answer