A block of mass m = 3.7 kg is on an inclined plane with a coefficient of friction μ1 = 0.24, at an initial height h = 0.64 m above the ground. The plane is inclined at an angle θ = 51°. The block is then compressed against a spring a distance Δx = 0.11 m from its equilibrium point (the spring has a spring constant of k1 = 31 N/m) and released. At the bottom of the inclined plane is a horizontal plane with a different coefficient of friction, μ2 = 0.055. After a distance d = 0.14 m, there is another spring (spring constant k2 = 6.9 N/m) under which the surface is frictionless. Use energy considerations in this problem.
(a) Assume the block does not make it all the way to the bottom of the ramp before it comes to rest. What is the x-component of the block's position when it stops? Use an x-axis parallel to the incline, with the positive direction pointing up the incline and the origin at the spring's equilibrium position. This should not depend on the initial height of the block, only on the angle.
(b) Let us now assume the values are such that the block does reach the bottom of the ramp. Find the speed of the block, in m/s, when it reaches the end of the ramp. Treat the block as a point.
(c) Once the block reaches the bottom, it encounters a new coefficient of friction. What distance x, in meters, could the block travel on this surface before stopping if no obstacles were in the way? Assume the size of the block is negligible in transitioning from the ramp to the plane.
(d) After a distance d on the surface with coefficient u2, the block encounters a frictionless surface and a spring with constant ky. Calculate how far the spring is compressed Δs in m when the block comes to rest.

a) 0.11825

b) 3.401 m/s

c)  10.72 m  

d) 2.47 m

Explanation:

Given:-

- The mass of the block, m = 3.7 kg

- The angle of the inclined plane, θ = 51°

- The spring constant of inclined spring, k1 = 31 N/m

- The spring constant of horizontal spring, k2 = 6.9 N/m

- The coefficient of friction ( inclined plane ) μ1 = 0.24

- The coefficient of friction ( horizontal surface ) μ2 = 0.055

- The initial compression of inclined spring, Δx = 0.11 m

- The initial height of the block, h = 0.64 m

- The horizontal distance to second spring, d = 0.14 m

a)

Solution:-

- We will denote the distance along the inclined surface as "Δs" that the block travels when released from its initial position: ( xi = Δx ).

- The final position where the block on the inclined plane comes to a stop is x = x1. The displacement Δs can be written as:

                         Δs = xi - x

                        Δs = Δx - x1

Note: The equilibrium position is considered as the origin.

The work-done by the block against friction ( W ):

                   W = F*Δs  

                   W = u*m*g*cos ( θ )*Δs

- Use a energy balance for the block between the initial compressed point and the final point on the inclined surface where the block comes to a stop:

                  U1 + Ep1 = Ep2 + W

Where,

          U1 : The elastic potential energy = 0.5*k1*Δx^2  

          Ep1: The initial gravitational potential energy = m*g*( h + Δx*sin ( θ ) )

          Ep2: The final gravitational potential energy = m*g*( h1 )

- The change in gravitational potential energy ΔEp = Ep2 - Ep1:

                  ΔEp = Ep2 - Ep1 = -m*g*( h + Δx*sin ( θ ) ) + m*g*( h1 )

                                              = m*g*( (h1 - h) - Δx*sin ( θ ) )

                                              = m*g*( -x1*sin ( θ ) - Δx*sin ( θ ) )

                                              = m*g*sin ( θ )* ( -x1 - Δx )

                                              = - m*g*sin ( θ )*( Δs )

- Use the energy principle expression stated above and solve for Δs:

              0.5*k1*Δx^2  = - m*g*sin ( θ )*( Δs )  + u1*m*g*cos ( θ )*Δs

              0.5*k1*Δx^2 = Δs [ m*g* ( u1*cos ( θ ) - sin ( θ ) ) ]

              0.5*31*0.11^2 = Δs [ 3.7*9.81* ( 0.24*cos ( 51° ) - sin ( 51° ) ) ]

              0.18755 = -22.72588*Δs

              Δs = - 0.00825 m  

The x-coordinate of the resting point would be:

              Δs = Δx - x1

              x1 = Δx - Δs

              x1 = 0.11 - ( -0.00825 )

              x1 = 0.11825 m

b)

Solution:-

Use a energy balance for the block between the initial compressed point and the final point at the bottom of inclined surface where the block has a velocity "u":              

                         U1 + Ep1 = W + Ek + Ep2

Where,

          U1 : The elastic potential energy = 0.5*k1*Δx^2  

          Ep1: The initial gravitational potential energy = m*g*( h + Δx*sin ( θ ) )

          Ek: The kinetic energy at the bottom of inclined surface = 0.5*m*u^2

Note: Taking the horizontal surface as the datum ( Ep2 = 0 )

Therefore,

  0.5*k1*Δx^2 + m*g*( h + Δx*sin ( θ ) ) = u1*m*g*cos ( θ )*Δs + 0.5*m*u^2                          

Where,

Δs: The total distance from initial point to bottom surface = Δx + h / sin ( θ )

 0.5*k1*Δx^2 + m*g*( h + Δx*sin ( θ ) ) = u1*m*g*cos ( θ )*(Δx + h / sin ( θ )) + 0.5*m*u^2  

0.5*31*0.11^2 + 3.7*9.81*( 0.64 + 0.11*sin ( 51° ) ) = 0.24*3.7*9.81*cos ( 51° )* ( 0.11 + 0.64 / sin ( 51° ) ) + 0.5*3.7*u^2

               0.18755 + 26.33296 = 5.11776 + 1.85*u^2

               21.40275 = 1.85*u^2

               u = √( 21.40275 / 1.85 ) = √11.56905

              u = 3.401 m/s   Answer

c)

Solution:-

- Use a energy balance for the block between the point at the bottom of the inclined surface and the final point where block reaches its maximum distance "s" by doing work against friction:

                                Ek = W1

                   0.5*m*u^2 = μ2*m*g*s

                     0.5*u^2 = μ2*g*s

             s = 0.5*3.401^2 / ( 0.055*9.81)

        s = 10.71893 m ≈ 10.72 m ... Answer

d)  

Solution:-

- The new friction force acting on the block acts for the distance of d = 0.14m.  

- The initial kinetic energy of the block corresponding to the speed ( u ) at the bottom of the inclined surface is reduced to speed ( v ) due to loss of kinetic energy by working against the friction.

- Apply the work-done principle against the new friction over the distance d travelled by the block on the horizontal surface is expressed as:

                          Ek1 = W2 + Ek2  

              0.5*m*u^2 = μ2*m*g*d + 0.5*m*v^2

- The final velocity of block ( v ) after doing work against the friction ( W2 ) can be determined by the above expressed energy principle:

                     u^2 - 2*μ2*g*d = v^2

                     v^2 = 3.40133^2 - 2*(0.055)*(9.81)*(0.14)

                     v = √11.41797

                     v = 3.37904 m/s

- After doing work against the friction the block's kinetic energy is stored into the spring in the form of elastic potential energy ( U2 ).  

- The conservation of energy principle can be applied ( No fictitious work done ).

                                Ek = U2

                    0.5*m*v^2 = 0.5*k2*Δs^2

                   Δs = v*√(m / k2) = 3.37904*√(3.7 /6.9) =  

                   Δs = 2.47440 m ≈ 2.47 m ... Answer

current and voltage can both be different in different parts of a circuit so it matters where you place the ammeter and voltmete