Question

A block with mass 5 kg is attached to the end of a horizontal spring with spring constant 200N/m. The other end of the spring is attached to a wall. The spring is stretched 10cm in the positive directions from its equilibrium length. Assume that the block is resting on a frictionless surface.

A) When the spring is fully stretched, what is the magnitude of the force from the spring on the block?

B) We then release the block, letting it slide in the negative direction. At the moment we release the block what is its acceleration?

C) As the block slides, does the magnitude of its acceleration increase, decrease, or stay the same?

D) When the block is released, how much kinetic energy does it have? How much potential energy does it have?

E) As the block slides, how does the total energy change? How does it kinetic energy change? How does its potential energy change?

F) At what point during its slide does the block have the greatest speed? What is that speed?

Answer 1

The block of mass m = 5 kg, the sping constant k = 200 N/m , the max displacement d= 10 cm

A) The fully stretched spring force exerted by the spring is Fs = - k d

k = 200 N/m and d= 10 cm = 0.1 m

so Fs = - 20 N or 20 N directed towards the equilibrium point

B) Let the accleration be a .

The force due to spring due to is equal to the force due to which the block is moving so

Fs = m. a

so Fs = 20 N and m = 5 kg

so , a = 20 / 5 = 4 m/s2 towards equilibrium point

C) as the block slides the force reduces upto eqilibrium point and becomes zero and then it again increase in the negative side

so the accleration first decreases to Zero upto equilibrium point and then it increases upto -10 cm

D) When the block was released the velocity was 0 so the Kinetic energy was also 0 , KE = 0

The potential energy PE is

where is the diaplacement from equilibrium point , and when released x= d= 10 cm

so PE = 1 J

E) As spring system obeys Law of Conservation of Energy ,

the total energy remains constant ,

The kinetic energy starts from 0 upto max value at equilibrium point   and it falls to 0 in the extreme point

The potential energy starts with max value and decreases to 0 at equilibrium point and then it increases again at the extreme points

F) The block has the greatest speed at the equilibrium point X= 0 .

the speed is given by    where A = amplitude = 10 cm

so

putting the values we get

V max = 0.632 m/s