In Fig. 6-50, block 1 of mass ${\mathbf{m}}_{\mathbf{1}}\mathbf{=}\mathbf{2}\mathbf{.}\mathbf{0}\mathbf{kg}$and block 2 of mass${\mathbf{m}}_{\mathbf{2}}\mathbf{}\mathbf{=}\mathbf{}\mathbf{3}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{kg}$are connected by a string of negligible mass and are initially held in place. Block 2 is on a frictionless surface tilted at$\mathbf{\theta }\mathbf{=}\mathbf{}\mathbf{30}\mathbf{°}\mathbf{}$. The coefficient of kinetic friction between block 1 and the horizontal surface is$\mathbf{0}\mathbf{.}\mathbf{25}\mathbf{}$. The pulley has negligible mass and friction. Once they are released, the blocks move. What then is the tension in the string?

• The mass of block 1,${\mathrm{m}}_1=2.0\text{\hspace{0.17em}kg}$.
• The mass of block 2,${\mathrm{m}}_2=3.0\text{\hspace{0.17em}kg}$.
• The tilted angle,$\theta =30°$.
• The kinetic friction coefficient between block 1 and the horizontal surface is 0.25.

The problem deals with Newton's second law of motion, which tells that the acceleration a, of an object can be determined by the net force F, acting upon the object and the mass m, of the object. Apply Newton's second law to each block's x-axis to solve the question.

Formula:

$\mathrm{F}=\mathrm{ma}$

Newton's second law on block 1:

$\mathrm{T}-{\mathrm{f}}_{\mathrm{k}}={\mathrm{m}}_1\mathrm{a}$

Newton's second law on block 2:

${\mathrm{m}}_2\mathrm{gsin}\left(\mathrm{\theta }\right)-\mathrm{T}={\mathrm{m}}_2\mathrm{a}$

Add the above two equations and calculate acceleration:

$\begin{array}{c}{\mathrm{m}}_1\mathrm{a}+{\mathrm{m}}_2\mathrm{a}={\mathrm{m}}_2\mathrm{gsin}\left(\mathrm{\theta }\right)-{\mathrm{f}}_{\mathrm{k}}\\ \mathrm{a}=\frac{{\mathrm{m}}_2\mathrm{gsin}\left(\mathrm{\theta }\right)-{\mathrm{\mu }}_{\mathrm{k}}{\mathrm{m}}_1\mathrm{g}}{{\mathrm{m}}_1+{\mathrm{m}}_2}\end{array}$

Substitute the values in the above expression, and we get,

$\begin{array}{l}\mathrm{a}=\frac{\left(3\text{\hspace{0.17em}kg}\right)\times \left(9.8\frac{\text{m}}{{\text{s}}^{\text{2}}}\right)\times \text{sin}(30°)-\left(0.25\right)\times \left(\text{2\hspace{0.17em}kg}\right)\times \left(9.8\frac{\text{m}}{{\text{s}}^{\text{2}}}\right)}{2\text{\hspace{0.17em}kg}+\text{3\hspace{0.17em}kg}}\\ \mathrm{a}=1.96\frac{\text{m}}{{\text{s}}^{\text{2}}}\end{array}$

Substitute the value of acceleration in the first equation to calculate tension as:

$\mathrm{T}={\mathrm{m}}_1\mathrm{a}+{\mathrm{f}}_{\mathrm{k}}$

Substitute the values in the above expression, and we get,

$\begin{array}{l}\mathrm{T}=\left(\text{2\hspace{0.17em}kg}\right)\times \left(\text{1.96\hspace{0.17em}}\frac{\text{m}}{{\text{s}}^{\text{2}}}\right)+\left(\text{0.25}\right)\times \left(\text{2\hspace{0.17em}kg}\right)\times \left(\text{9.8\hspace{0.17em}}\frac{\text{m}}{{\text{s}}^{\text{2}}}\right)\\ \mathrm{T}=8.8\text{\hspace{0.17em}N}\end{array}$

Thus, the tension in the string $8.8\text{\hspace{0.17em}N}$.