Chapter 11, Problem 71A

a.

To determine

Speed ofK.

Answer to Problem 71A

3.429 m/s

Explanation of Solution

Given:

Weight of K, $W=420\text{ N}$

Initial height, $h_1=0.4\text{ m}$

Final height, $h_2=1\text{ m}$

Formula used:

Change in potential energy is given by

  $\Delta P=mg\left(h_2-h_1\right)$

Here, m is mass, g is acceleration due to gravity and h is height.

Kinetic energy is given by

  $KE=\frac{1}{2}m{v}^2$

Here, m is mass and vis velocity.

According to energy conservation.

  $mg\left(h_2-h_1\right)=\frac{1}{2}m{v}^2$

Calculation:

Now, substitute the value of m, g, $h_2$ and $h_1$ and solve.

  $\begin{array}{l}mg\left(h_2-h_1\right)=\frac{1}{2}m{v}^2\\ g\left(h_2-h_1\right)=\frac{1}{2}{v}^2\\ 9.8\left(1-0.4\right)=\frac{1}{2}{v}^2\\ v^2=11.76\\ v=3.429\text{ m/s}\end{array}$

Conclusion:

Therefore,K is moving with speed of 3.429 m/s.

b

To determine

Work done by friction.

Answer to Problem 71A

  $W=-85.714\text{ J}$

Explanation of Solution

Given:

Weight of K, $w=420\text{ N}$

Initial height, $h_1=0.4\text{ m}$

Final height, $h_2=1\text{ m}$

Speed of K at lowest point is, $u=2\text{ m/s}$

Formula used:

According to work energy theorem work done is equal to change in kinetic energy.

  $W=\frac{1}{2}m\left(v^2-u^2\right)$........(1)

Weight:

  $w=mg$........(2)

Here, m is mass and gis acceleration due to gravity.

Calculation:

Now, substitute the value of w and g in equation 1 and solve.

  $\begin{array}{l}420=m\times 9.8\\ m=\frac{420}{9.8}\\ =42.857\text{ kg}\end{array}$

Due to frictional force,K will finally stop.

Therefore, $v=0\text{ m/s}$

Now, substituting value of m, v and u in equation 2 and solve.

  $\begin{array}{l}W=\frac{1}{2}42.857\left(0^2-2^2\right)\\ =\frac{1}{2}\times 42.857\times -4\\ =-85.714\text{ J}\end{array}$

Here, negative sign indicates that work is done by friction.

Conclusion:

Therefore, work done by friction is 85.714 J.