Chapter 6, Problem 65A

A 10-kg mass on a horizontal friction-free air track is accelerated by a string attached to another 10-kg mass hanging vertically from a pulley as shown. What is the force due to gravity, in newtons, on the hanging 10-kg mass? What is the acceleration of the system of both masses?

To determine

The force due to gravity on the hanging mass and the acceleration of the system of both masses.

Answer to Problem 65A

The force due to gravity on the hanging mass is $98\text{ N}$ .

The acceleration of the system of both masses is $a=4.9\text{m}/{\text{s}}^{\text{2}}$ .

Explanation of Solution

Given info:

The mass on the horizontal friction-free air track is $m_1=10\text{ kg}$ .

The vertical hanging mass is $m_2=10\text{ kg}$ .

Formula used:

The expression of weight as follows:

  $W=mg$

Here, $m$ is the mass and $g$ is the acceleration due to gravity.

The expression for Newton’s second law of acceleration as follows:

  $F=ma$

Here, $m$ is the mass and $a$ is the acceleration.

Calculation:

Consider the acceleration due to gravity as $g=9.8\text{m}/{\text{s}}^{\text{2}}$ .

The force due to gravity on the hanging mass as follows:

  $\begin{array}{l}W=m_{2}g\\ F_{\text{net}}=\left(10\text{ kg}\right)\times \left(9.8\text{m}/{\text{s}}^{\text{2}}\right)\\ F_{\text{net}}=98\text{ N}\end{array}$

Hence, the force due to gravity on the hanging mass is $98\text{ N}$ .

The total mass on the system as follows:

  $\begin{array}{l}m=m_1+m_2\\ m=10\text{ kg}+10\text{ kg}\\ m=20\text{ kg}\end{array}$

The acceleration of the system of both masses as follows:

  $\begin{array}{l}F=ma\\ a=\frac{F_{\text{net}}}{m}\\ a=\frac{98\text{ N}}{20\text{ kg}}\\ a=4.9\text{m}/{\text{s}}^{\text{2}}\end{array}$

Hence, the acceleration of the system of both masses is $a=4.9\text{m}/{\text{s}}^{\text{2}}$ .

Conclusion:

Thus, the force due to gravity on the hanging mass is $98\text{ N}$ .

Thus, the acceleration of the system of both masses is $a=4.9\text{m}/{\text{s}}^{\text{2}}$ .