A 2.0 kg mass is pushed against a spring and compresses it a distance of 0.50 m. The spring constant is 104 N/m and the frictional force between the mass and the horizontal surface on which it lies is constant an its magnitude is 8.0 N. When the mass is released, calculate the speed with which it will leave the spring.

Question

anonymous · Answer

total force acting at the block is F= kx-8 =ma= m(dv/dt)= mv(dv/dx)
so mv dv = (kx-8)
$$\int\limits_{0}^{v} mv dv=\int\limits_{0.5}^{0} (kx-8)dx$$
you know values of k and m 
i hope this will help you.

anonymous · Answer

I don't know what that symbol is, we haven't learned anything like that :c

anonymous · Answer

Lol..

anonymous · Answer

This can be done using the concept of energy mate!.. have you learnt about spring potential energy?

anonymous · Answer

Yes but I don't understand it :/

anonymous · Answer

what part don't you get ?
when u compress a spring.. you have to do work.. and so you have to spend energy (this will your chemical energy which you acquired via breathing and food) and this energy gets stored up in the compressed spring as spring potential energy

The expression for it.. $$U = \frac{kx^2}{2}$$.. where x is the displacement from the mean position (ur springs undisturbed resting position)
get it?

anonymous · Answer

Thanks I got it actually nevermind