A single mass (m1 = 3.5 kg) hangs from a spring in a motionless elevator. The spring constant is k = 278 N/m. 1)What is the distance the spring is stretched from its unstretched length?

0.126m

According to Hooke's law, the force (F) acting on a spring to cause an extension or compression (e) is given by;

F = k x e            (i)

Where;

k = the spring's constant.

From the question, the force acting on the spring is the weight(W) of the mass. i.e

F = W               (ii)

But;

W = m x g;

where;

m = mass of the object

g = acceleration due to gravity [usually taken as 10m/s²]

From equation (ii), it implies that;

F = W = m x g

Now substitute F = m x g into equation(i) as follows;

F = k x e

m x g = k x e      (iii)

From the question;

m = m1 = 3.5kg

k = 278N/m

Substitute these values into equation (iii) as follows;

3.5 x 10 = 278 x e

35 = 278e

Now solve for e;

e = 35/278

e = 0.126m

Therefore, the distance the spring is stretched from its unstretched length (which is the same as the extension of the spring) is 0.126m

  distance on the y-axis and time on the x-axis.

explanation:

the instantaneous velocity = change in position/time taken

the slope of the graph should give instantaneous velocity.

  slope of a graph = change in value of y -axis / change in values of x -axis

  comparing both the equations

    change in value of y -axis = change in position

  change in values of x -axis = time taken

so position values should be on the y axis and time values should be on the x axis.

distance on the y-axis and time on the x-axis.

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.