Question 8.8: Block–Spring Collision A block having a mass of 0.80 kg is g......

(A) Conceptualize The various parts of Figure 8.11 help us imagine what the block will do in this situation. All motion takes place in a horizontal plane, so we do not need to consider changes in gravitational potential energy.

Categorize We identify the system to be the block and the spring. The block-spring system is isolated with no nonconservative forces acting.

Analyze Before the collision, when the block is at Ⓐ, it has kinetic energy and the spring is uncompressed, so the elastic potential energy stored in the system is zero. Therefore, the total mechanical energy of the system before the collision is just \frac{1}{2} m v_{Ⓐ}{ }^{2}. After the collision, when the block is at Ⓒ, the spring is fully compressed; now the block is at rest and so has zero kinetic energy. The elastic potential energy stored in the system, however, has its maximum value \frac{1}{2} k x^{2}=\frac{1}{2} k x_{\max }^{2}, where the origin of coordinates x=0 is chosen to be the equilibrium position of the spring and x_{\max } is the maximum compression of the spring, which in this case happens to be x_{Ⓒ}. The total mechanical energy of the system is conserved because no nonconservative forces act on objects within the isolated system.

Write a conservation of mechanical energy equation:

\begin{aligned} & K_{Ⓒ}+U_{s Ⓒ}=K_{Ⓐ}+U_{s Ⓐ} \\ & 0+\frac{1}{2} k x_{\max }^{2}=\frac{1}{2} m v_{Ⓐ}^{2}+0 \end{aligned}

Solve for x_{\max } and evaluate:

x_{\max }=\sqrt{\frac{m}{k}} v_{Ⓐ}=\sqrt{\frac{0.80 \mathrm{~kg}}{50 \mathrm{~N} / \mathrm{m}}}(1.2 \mathrm{~m} / \mathrm{s})=0.15 \mathrm{~m}

(B) Conceptualize Because of the friction force, we expect the compression of the spring to be smaller than in part (A) because some of the block’s kinetic energy is transformed to internal energy in the block and the surface.

Categorize We identify the system as the block, the surface, and the spring. This system is isolated but now involves a nonconservative force.

Analyze In this case, the mechanical energy E_{\text {mech }}=K+U_{s} of the system is not conserved because a friction force acts on the block. From the particle in equilibrium model in the vertical direction, we see that n=m g.

Evaluate the magnitude of the friction force:

f_{k}=\mu_{k} n=\mu_{k} m g

Write the change in the mechanical energy of the system due to friction as the block is displaced from x=0 to x_{Ⓒ} :

\Delta E_{\text {mech }}=-f_{k} x_{Ⓒ}

Substitute the initial and final energies:

\Delta E_{\text {mech }}=E_{f}-E_{i}=\left(0+\frac{1}{2} k x_{Ⓒ}^{2}\right)-\left(\frac{1}{2} m v_{Ⓐ}^{2}+0\right)=-f_{k} x_{Ⓒ}

\frac{1}{2} k x_{Ⓒ}^{2}-\frac{1}{2} m v_{Ⓐ}^{2}=-\mu_{k} m g x_{Ⓒ}

Substitute numerical values:

Solving the quadratic equation for x_Ⓒ gives x_Ⓒ = 0.093 m and x_Ⓒ = -0.25 m. The physically meaningful root is x_Ⓒ = 0.093 m.

Finalize The negative root does not apply to this situation because the block must be to the right of the origin (positive value of x ) when it comes to rest. Notice that the value of 0.093 \mathrm{~m} is less than the distance obtained in the frictionless case of part (A) as we expected.