A small block with a mass of 0.0600 kg is attached to a cord passing through a hole in a frictionless, horizontal surface (Fig. P6.71). The block is originally revolving at a distance of 0.40 m from the hole with a speed of 0.70 m/s . The cord is then pulled from below, shortening the radius of the circle in which the block revolves to 10 m . At this new distance, the speed of the block is 2.80 m/s. (a) What is the tension in the cord in the original situation, when the block has speed $\mathbf{v}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{70}\mathbf{}\mathbf{m}\mathbf{/}\mathbf{s}$? (b) What is the tension in the cord in the final situation, when the block has speed $\mathbf{v}\mathbf{=}\mathbf{2}\mathbf{.}\mathbf{80}\mathbf{}\mathbf{m}\mathbf{/}\mathbf{s}$?

The given data can be listed below,

• The mass of the block is,$m=0.060\mathrm{kg}$
• The initial speed of block is,$v_j=0.70\mathrm{m}/{\mathrm{s}}^2$
• The initial distance at which block is revolving,$r_j=0.4\mathrm{m}$
• The final speed of the block is,$v_f=2.80\mathrm{m}/\mathrm{s}$

When an item is pulled from both sides, a force known as tension acts axially. Being a force, it frequently has units like pounds or newtons.

The tension in the cord is equal to the centripetal force which can be given by,

$T_i=\frac{m{v}_i^2}{r_i}$

Here, m is the mass of the block,$v_i$ is the initial velocity of the block, and$r_i$ is the initial distance at which block is revolving.

Substitute all the values in the above,

$$\begin{gathered} T=\frac{\left(0.06\mathrm{kg}\right){\left(0.7\mathrm{m}/{\mathrm{s}}^2\right)}^2}{0.4\mathrm{m}} \\ =0.07\mathrm{N} \end{gathered}$$

Thus, the tension in the string when block is revolving with a velocity 0.70 m/s is 0.07 N

The tension in the cord is equal to the centripetal force which can be given by,

$T_f=\frac{m{v}_f^2}{r_f}$

Here, m is the mass of the block,$v_f$ is the final velocity of the block, and$r_f$ is the final distance at which block is revolving.

Substitute all the values in the above,

$$\begin{gathered} T=\frac{\left(0.06\mathrm{kg}\right){\left(2.80\mathrm{m}/{\mathrm{s}}^2\right)}^2}{0.10\mathrm{m}} \\ =4.70\mathrm{N} \end{gathered}$$

Thus, the tension in the string when block is revolving with a velocity 2.80 m/s is 4.70 N