Ablock with mass m = 7.4 kg is attached to two springs with spring constants kleft = 31 n/m and kright = 53 n/m. the block is pulled a distance x = 0.27 m to the left of its equilibrium position and released from rest.

1)what is the magnitude of the net force on the block (the moment it is released)?

n

2)what is the effective spring constant of the two springs?

n/m

3)what is the period of oscillation of the block?

s

4)how long does it take the block to return to equilibrium for the first time?

s

5)what is the speed of the block as it passes through the equilibrium position?

m/s

6)what is the magnitude of the acceleration of the block as it passes through equilibrium?

m/s2

7)where is the block located, relative to equilibrium, at a time 1.06 s after it is released? (if the block is left of equilibrium give the answer as a negative value; if the block is right of equilibrium give the answer as a positive value)

m

8)what is the net force on the block at this time 1.06 s? (a negative force is to the left; a positive force is to the right)

n

9)what is the total energy stored in the system?

j

10)if the block had been given an initial push, how would the period of oscillation change?

the period would increase

the period would decrease

the period would not change

i need with this question

Explanation:

1) Net force: According to Newtons second law, when a net force is acting on an object, the object must be accelerating i.e its speed changes from second to second. Also, Net force is the sum of all forces acting on an object. If the object is not accelerating the sum of net forces is zero

mathematically, Fnet = 0

∑Fₓ = 0, let the constants be represented as k

x = mass of the block

∴ F - K1x - K2x = 0

F = K1x + K2x

F = x(K1 + K2)

  = 0.27(31 + 53)

  = 22.63N

2) Effective spring constant

K_eff = K1 + K2

         = 31 + 53

         = 84N/m

3) Period of Oscillation of the block (T in second)

  ω = angular speed/frequency

  ω = √k_eff/mass of block

      = √84/7.4

     ω = √11.351 = 3.369rad/s

Also ω = 2π/T

∴T = 2π/ω

     = 2π/3.369

   T =1.86s

4) How long does it take the block to return to equilibrium for the first time?

period is 1.86s, from 0.27m to 0 is a quarter of one cycle,

so t = period/4

t = 1.86/4

 = 0.465s

5) What is the speed of the block as it passes through the equilibrium position?

To determine the speed of the block, speed is maximum at equilibrium point (V = velocity or speed)

∴Vmax = mass of the block/angular speed

            = 0.27/3.369

 Vmax = 0.08m/s

6) What is the magnitude of the acceleration of the block as it passes through equilibrium?

The object is not accelerating, the object is at a particular position

Fnet = 0

In other words, at equilibrium acceleration is zero. magnitude of acceleration is zero.

7)Where is the block located, relative to equilibrium, at a time 1.06 s after it is released?

x(t) = Acos(ωt + φ)

x(t) = 0.27cos(3.369t + φ)

-0.27 = 0.27cos(3.369(0) + φ)

    -1 = cosφ

φ = π

To calculate where the block is located

x(t) = Acos(ωt + φ)

x(t) = 0.27cos(3.369t + π)

x(1.06) = 0.27cos(3.369*1.06 +π)

x(0.08) = 0.245m (right)

explanation:

newton's third law, for every action there is an equal and opposite reaction.

an energy transformation flow diagram is shown

where is it ?