Chapter 7, Problem 69P

(a)

To determine

The distance traveled by block along the incline plane.

Explanation of Solution

Given:

The mass of block is $2.4\text{kg}$ .

The initial speed of block is $3.8\text{m}/\text{s}$ .

The coefficient of friction is $0.30$ .

The angle of incline with the horizontal is $37°$ .

Formula used:

Write the expression for friction force along the incline.

  $f=\mu mg\cos \theta$

Here, $m$ is mass of block, $g$ is gravitational acceleration, $\mu$ is the coefficient of friction, $f$ is the friction force and $\theta$ is the angle of weight of block with the perpendicular to the incline.

Write the expression for thermal energy.

  $\Delta E_{\text{thermal}}=f\Delta s$

Here, $\Delta E_{\text{thermal}}$ is the change in thermal energy and $\Delta s$ is the displacement covered by block.

Substitute $\mu mg\cos \theta$ for $f$ in above expression.

  $\Delta E_{\text{thermal}}=\left(\mu mg\cos \theta \right)\Delta s$

Write the expression for change in potential energy.

  $\Delta U=mg\left(h_f-h_i\right)$

Here, $\Delta U$ is the potential energy stored in the body, $h_i$ is the initial height of block and $h_f$ is the final height of block.

Write the expression for change in kinetic energy.

  $\Delta K=\frac{1}{2}m\left(v_f^2-v_i^2\right)$

Here, $\Delta K$ is the change in kinetic energy, $v_i$ is the initial velocity of block and $v_f$ is the final velocity of block.

Total energy of block is conserved at all points. Work done by external force is equal to the sum of change in gravitational potential energy, kinetic energy and thermal energy.

Write the expression of work done by external force.

  $W_{\text{external}}=\Delta U+\Delta K+\Delta E_{\text{thermal}}$ ........(1)

Here, $W_{\text{external}}$ is the work done by external force.

  

There is no external force acting on the block; so, the work done by external force is zero.

Substitute $\frac{1}{2}m\left(v_f^2-v_i^2\right)$ for $\Delta K$ , $\left(\mu mg\cos \theta \right)\Delta s$ for $\Delta E_{\text{thermal}}$ , $0$ for $W_{\text{external}}$ and $mg\left(h_f-h_i\right)$ for $\Delta U$ in equation (1).

  $0=\frac{1}{2}m\left(v_f^2-v_i^2\right)+mg\left(h_f-h_i\right)+\left(\mu mg\cos \theta \right)\Delta s$

Substitute $\Delta s\sin \theta$ for $h_f$ , $0$ for $h_i$ , $0$ for $v_f$ and rearrange the above expression in terms of $\Delta s$ .

  $\Delta s=\frac{v_i^2}{2g\left(\sin \theta +\mu \cos \theta \right)}$ ........(2)

Calculation:

Substitute $3.8\text{m}/\text{s}$ for $v_i$ , $0.30$ for $\mu$ , $37°$ for $\theta$ and $9.81\text{m}/{\text{s}}^2$ for $g$ in equation (2).

  $\begin{array}{c}\Delta s=\frac{{\left(3.8\text{m}/\text{s}\right)}^2}{2\left(9.81\text{m}/{\text{s}}^2\right)\left(\sin 37°+\left(0.30\right)\cos 37°\right)}\\ =0.87\text{m}\end{array}$

Conclusion:

Thus, the distance traveled by block along the incline is $0.87\text{m}$ .

(b)

To determine

The speed of the block when it has traveled half the distance in part (a).

Explanation of Solution

Given:

The mass of block is $2.4\text{kg}$ .

The initial speed of block is $3.8\text{m}/\text{s}$ .

The coefficient of friction is $0.30$ .

The angle of incline with the horizontal is $37°$ .

Formula used:

Write the expression for friction force along the incline.

  $f=\mu mg\cos \theta$

Here, $m$ is mass of block, $g$ is gravitational acceleration, $\mu$ is the coefficient of friction, $f$ is the friction force and $\theta$ is the angle of weight of block with the perpendicular to the incline.

Write the expression for thermal energy.

  $\Delta E_{\text{thermal}}=f\Delta s$

Here, $\Delta E_{\text{thermal}}$ is the change in thermal energy and $\Delta s$ is the displacement covered by block.

Substitute $\mu mg\cos \theta$ for $f$ in above expression.

  $\Delta E_{\text{thermal}}=\left(\mu mg\cos \theta \right)\Delta s$

Write the expression for change in potential energy.

  $\Delta U=mg\left(h_f-h_i\right)$

Here, $\Delta U$ is the potential energy stored in the body, $h_i$ is the initial height of block and $h_f$ is the final height of block.

Write the expression for change in kinetic energy.

  $\Delta K=\frac{1}{2}m\left(v_f^2-v_i^2\right)$

Here, $\Delta K$ is the change in kinetic energy, $v_i$ is the initial velocity of block and $v_f$ is the final velocity of block.

Total energy of block is conserved at all points. Work done by external force is equal to the sum of change in gravitational potential energy, kinetic energy and thermal energy.

Write the expression of work done by external force.

  $W_{\text{external}}=\Delta U+\Delta K+\Delta E_{\text{thermal}}$

Here, $W_{\text{external}}$ is the work done by external force.

  

There is no external force acting on the block; so, the work done by external force is zero.

Substitute $\frac{1}{2}m\left(v_f^2-v_i^2\right)$ for $\Delta K$ , $\left(\mu mg\cos \theta \right)\Delta s$ for $\Delta E_{\text{thermal}}$ , $0$ for $W_{\text{external}}$ and $mg\left(h_f-h_i\right)$ for $\Delta U$ in equation (1).

  $0=\frac{1}{2}m\left(v_f^2-v_i^2\right)+mg\left(h_f-h_i\right)+\left(\mu mg\cos \theta \right)\Delta s$

Substitute $\Delta s\sin \theta$ for $h_f$ , $0$ for $h_i$ , $0$ for $v_f$ and rearrange the above expression in terms of $\Delta s$ .

  $\Delta s=\frac{v_i^2}{2g\left(\sin \theta +\mu \cos \theta \right)}$

For object velocity at halfway on incline plane:

Substitute $\frac{1}{2}m\left(v_f^2-v_i^2\right)$ for $\Delta K$ , $\left(\mu mg\cos \theta \right)\frac{\Delta s}{2}$ for $\Delta E_{\text{thermal}}$ , $0$ for $W_{\text{external}}$ and $mg\left(h_f-h_i\right)$ for $\Delta U$ in equation (1).

  $0=\frac{1}{2}m\left(v_f^2-v_i^2\right)+mg\left(h_f-h_i\right)+\left(\mu mg\cos \theta \right)\frac{\Delta s}{2}$

Substitute $\left(\frac{\Delta s}{2}\right)\sin \theta$ for $h_f$ , $0$ for $h_i$ and rearrange the above expression in terms of $v_f$ .

  $v_f=\sqrt{v_i^2-g\Delta s\left(\sin \theta +\mu \cos \theta \right)}$ ........(3)

Calculation:

Substitute $3.8\text{m}/\text{s}$ for $v_i$ , $0.30$ for $\mu$ , $0.87\text{m}$ for $\Delta s$ , $37°$ for $\theta$ and $9.81\text{m}/{\text{s}}^2$ for $g$ in equation (3).

  $\begin{array}{c}{v}_f=\sqrt{{\left(3.8\text{m}/\text{s}\right)}^2-\left(9.81\text{m}/{\text{s}}^2\right)\left(0.87\text{m}\right)\left[\sin 37°+\left(0.30\right)\cos 37°\right]}\\ =2.7\text{m}/\text{s}\end{array}$

Conclusion:

Thus, the speed of the block when it has traveled half the distance in part (a)is $2.7\text{m}/\text{s}$ .