Question

A block of mass ${\mathbf{m}}_{\mathbf{1}}\mathbf{=}\mathbf{20}\mathbf{.0}\mathbf{}\mathbf{}\mathbf{kg}$is connected to a block of mass ${\mathbf{m}}_{\mathbf{2}}\mathbf{=}\mathbf{30.0}\mathbf{}\mathbf{}\mathbf{kg}$by a massless string that passes over a light, frictionless pulley. The $\mathbf{30.0}\mathbf{}\mathbf{}\mathbf{kg}$ block is connected to a spring that has negligible mass and a force constant of $\mathbf{k}\mathbf{=}\mathbf{250.0}\mathbf{}\mathbf{}\mathbf{N}\mathbf{/}\mathbf{m}$ as shown in Figure P8.64. The spring is unstretched when the system is as shown in the figure, and the incline is frictionless. The $\mathbf{20.0}\mathbf{}\mathbf{}\mathbf{kg}$ block is pulled a distance $\mathbf{h}\mathbf{=}\mathbf{20.0}\mathbf{}\mathbf{}\mathbf{m}$ down the incline of angle $\mathbf{\theta }\mathbf{=}\mathbf{40}\mathbf{°}$ and released from rest. Find the speed of each block when the spring is again unstretched.

Short answer

When the spring is unstretched, each block will move with the velocity $\mathrm{v}=1.24\text{m/s.}$

Step-by-step solution

Concept

In the absence of dissipative forces, the sum of the kinetic energy and the potential energy of a body remain constant. Thus, change is a mechanical energy of the system, and is zero.

$\mathbf{\Delta K}\mathbf{+}\mathbf{\Delta U}\mathbf{=}\mathbf{0}$

Step 2:

As no dissipative forces act on the system, using the law of conservation of energy, we get:

$\begin{array}{c}\mathrm{\Delta K}+\mathrm{\Delta U}=0\\ {(\mathrm{K}+\mathrm{U})}_{\mathrm{i}}={(\mathrm{K}+\mathrm{U})}_{\mathrm{f}}\\ 0+\left[\left({\mathrm{mgy}}_{\mathrm{i}}\right)+\frac{1}{2}{\mathrm{kx}}_{\mathrm{i}}^2\right]=\frac{1}{2}{\mathrm{mv}}_{\mathrm{f}}^2+{\mathrm{mgy}}_{\mathrm{f}}\end{array}$

Here, the subscript i denotes the values of the different quantities at the time of release of the block A, and subscript f denotes the values of the different quantities at the time when the spring is again unstretched.

For the given values, the above equation becomes as shown below.

$\begin{array}{c}\left(\begin{array}{l}0+(30.0\text{kg})\left({\text{9.8m/s}}^{\text{2}}\right)(0.200\text{m})+\frac{1}{2}\left(\text{250N/m}\right){(0.200\text{m})}^2\\ =\frac{1}{2}(20.0\text{kg}+30.0\text{kg}){\mathrm{v}}^2+(20.0\text{kg})(9.80{\text{m/s}}^{\text{2}})(0.200\text{m})\sin {40}^{\mathrm{o}}\end{array}\right)\\ 58.8\text{J}+5.00\text{J}=(25.0\text{kg}){\mathrm{v}}^2+25.2\text{J}\\ \mathrm{v}=\sqrt{\frac{33.6\text{J}}{25.0\text{kg}}}\\ =1.24\text{m/s.}\end{array}$

Thus, each block moves with a speed of: $\mathrm{v}=1.24\text{m/s}$.