Amass is attached to a spring, which is attached to a wall. the distance from the mass and the equilibrium distance, x0 = 0, is given by x. the spring constant is 5n/m. the equilibrium distance is 1m from the wall. i. what is the force exerted on the mass at x = 3m?
ii. what is the force exerted on the mass when the mass is touching the wall?
iii. what work must be done on the mass to move it from the wall to x = 3m?

Explanation:

Spring constant k = 5N/m

I ) x = 3 m means , spring is stretched by 3 m  

Restoring force by spring = kx = 5 x 3 = 15 N

II )  When mass is touching the wall, extension in spring = 1 m

Force by spring on the body

= 1 x 5 = 5 N .

iii ) . It is touching the wall , x =  1 m

Stored energy in the spring = 1/2 k x² = .5 x 5 x 1 x 1

= 2.5 J

When x = 3 , energy stored in it

Potential energy stored in it = 1/2 k x²

= .5 x 5 x 3 x 3

= 22.5 J

Increase in stored energy = 22.5 - 2.5

20 J

This must be the work done to stretch it from 1 m to 3 m .

when two balls of the different masses are dropped from equal height, they reach the ground at the same time .this is because they experience the same acceleration.

the acceleration is independent of mass.

thus when they are dropped they will always maintain the same velocity and travel the same distance (neglect air resistance of course) at the same time.