Question

In Fig. 9-57 a stationary block explodes into two pieces Land Rthat slide across a frictionless floor and then into regions with friction, where they stop. Piece L, with a mass of 2.0 kgencounters a coefficient of kinetic friction $\mathit{\mu }\mathit{L}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{40}$and slides to a stop in distance ${\mathit{d}}_{\mathbf{L}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{15}\mathbf{}\mathit{m}$. Piece Rencounters a coefficient of kinetic friction ${\mathit{d}}_{\mathbf{R}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{50}\mathbf{}\mathit{m}$and slides to a stop in distance ${\mathit{d}}_{\mathbf{R}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{25}\mathbf{}\mathit{m}$.What was the mass of the block?

Short answer

The mass of the block is,${\mathrm{m}}_{\mathrm{R}}+{\mathrm{m}}_{\mathrm{L}}\approx 3.4\mathrm{kg}$

Step-by-step solution

 Step 1: Given Data

The mass of piece L is,${\mathrm{m}}_{\mathrm{L}}=2.0\mathrm{kg}$

The coefficient of kinetic friction is${\mathrm{\mu }}_{\mathrm{L}}=0.40$ .

The coefficient of kinetic friction is${\mathrm{\mu }}_{\mathrm{R}}=0.50$ .

The distance is${\mathrm{d}}_{\mathrm{L}}=0.15\mathrm{m}$ .

The distance is${\mathrm{d}}_{\mathrm{R}}=0.25\mathrm{m}$ .

Determining the concept

By using the fact that kinetic energy gets converted to thermal energy and using the formula for frictional force, find the equation for square of velocity. From this equation of, find themass of the block.

Formulae are as follow:

The kinetic energy, $KE=\frac{1}{2}m{v}^2......\left(1\right)$

Frictional force f is,$f=\mu F_N.......\left(2\right)$

where, m is mass, v is velocity, f is frictional force,$F_N$is normal force,$\mu$is coefficient of friction and KE is kinetic energy.

Determining the mass of the block

Note that kinetic energy is getting converted to thermal energy. Thermal energy is generated because of the work done by the frictional force. Therefore, by equating these equations,

$\frac{1}{2}m{v}^2=f_{k}d$,

$f=\mu F_N\mathrm{and}{\mathrm{F}}_{\mathrm{N}}=\mathrm{mg}$,

Now, substituting,

$v^2=2\mu gd$

Thus, a ratio can be written as,

$$\begin{gathered} {\left(\frac{v_L}{v_R}\right)}^2=\frac{2{\mu }_{L}g{d}_L}{2{\mu }_{R}g{d}_R} \\ {\left(\frac{v_L}{v_R}\right)}^2=\frac{2\times 0.40\times 9.8\times 0.15}{2\times 0.50\times 9.8\times 0.25} \\ {\left(\frac{v_L}{v_R}\right)}^2=\frac{12}{25} \end{gathered}$$

But, the ratio of speed must be inversely proportional to the ratio of masses. Consequently,

${\left(\frac{m_R}{m_L}\right)}^2=\frac{12}{25}$

$$\begin{gathered} \frac{{\mathrm{m}}_{\mathrm{R}}}{{\mathrm{m}}_{\mathrm{L}}}=\sqrt{\frac{12}{25}} \\ {\mathrm{m}}_{\mathrm{R}}=\frac{2}{5}\sqrt{3}{\mathrm{m}}_{\mathrm{L}} \\ {\mathrm{m}}_{\mathrm{R}}=\frac{2}{5}\sqrt{3}\times 2 \\ {\mathrm{m}}_{\mathrm{R}}=1.39\mathrm{kg} \end{gathered}$$

Therefore, the total mass is,

$$\begin{gathered} {\mathrm{m}}_{\mathrm{R}}+{\mathrm{m}}_{\mathrm{L}}=1.39+2.0=3.39\mathrm{kg} \\ {\mathrm{m}}_{\mathrm{R}}+{\mathrm{m}}_{\mathrm{L}}\approx 3.4\mathrm{kg} \end{gathered}$$

Hence, the mass of the block is 3.4 kg.

Equating kinetic energy to the work done by the frictional force, the equation for the velocity can be found, from which, the mass can be found.