Chapter 11, Problem 56A

(a)

To determine

The change in kinetic energy of the car.

Answer to Problem 56A

  $\Delta KE=-1.44\times {10}^5\text{ J}$

Explanation of Solution

Given:

Mass of the car is $m=2\times {10}^3\text{ kg}$

Initial velocity of the car is $v_i=12.0\text{ m/s}$

Final velocity of the car is $v_f=0.0\text{ m/s}$

Formula used:

Kinetic energy $\left(KE\right)$ of a body of mass $\left(m\right)$ moving with velocity $\left(v\right)$ is given by,

  $KE=\frac{1}{2}m{v}^2$

Change in kinetic energy = final kinetic energy − initial kinetic energy

That is, $\Delta KE=K{E}_f-K{E}_i$

Calculation:

The change in kinetic energy of the car can be calculated by using the equation,

  $\Delta KE=\frac{1}{2}m{v}_f^2-\frac{1}{2}m{v}_i^2=\frac{1}{2}m\left(v_f^2-v_i^2\right)$  $\left(1\right)$

Substituting the numerical values in equation $\left(1\right)$ ,

  $\begin{array}{c}\Delta KE=\frac{1}{2}\left(2\times {10}^3\text{ kg}\right)\left[{\left(0.0\text{ m/s}\right)}^2-{\left(12.0\text{ m/s}\right)}^2\right]\\ =\left(1\times {10}^3\text{kg}\right)\left(-144{\text{ m}}^{\text{2}}{\text{/s}}^{\text{2}}\right)\\ =-1.44\times {10}^5\text{ kg}{\text{.m}}^{\text{2}}{\text{/s}}^{\text{2}}\\ =-1.44\times {10}^5\text{J}\end{array}$

Conclusion:

The change in kinetic energy of the car is $-1.44\times {10}^5\text{ J}$ .

(b)

To determine

The amount of work done by the tree on the car.

Answer to Problem 56A

  $W=-1.44\times {10}^5\text{ J}$

Explanation of Solution

Given:

Mass of the car is $m=2\times {10}^3\text{ kg}$

Initial velocity of the car is $v_i=12.0\text{ m/s}$

Final velocity of the car is $v_f=0.0\text{ m/s}$

The front of the car crashes in to the tree

Formula used:

Work done by the tree on the car can be calculated as,

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

Calculation:

Substituting the numerical values in equation $\left(2\right)$ ,

  $\begin{array}{c}W=\frac{1}{2}\left(2\times {10}^3\text{ kg}\right)\left[{\left(0.0\text{ m/s}\right)}^2-{\left(12.0\text{ m/s}\right)}^2\right]\\ =\left(1\times {10}^3\text{kg}\right)\left(-144{\text{ m}}^{\text{2}}{\text{/s}}^{\text{2}}\right)\\ =-1.44\times {10}^5\text{ kg}{\text{.m}}^{\text{2}}{\text{/s}}^{\text{2}}\\ =-1.44\times {10}^5\text{J}\end{array}$

Conclusion:

The amount of work done by the tree on the car as the front of the car crashes into the tree is $-1.44\times {10}^5\text{ J}$ .

(c)

To determine

The magnitude of the force.

Answer to Problem 56A

  $\left|F\right|=\text{2}\text{.88}\times {10}^5\text{ N}$

Explanation of Solution

Given:

The length through which the front of the car is pushed in by the force is $d\text{ = 50}\text{.0 cm = 0}\text{.5 m}$

From the part (b), the amount of work done by the tree on the car as the front of the car crashes in to the tree is $W=-1.44\times {10}^5\text{ kg}{\text{.m}}^{\text{2}}{\text{/s}}^{\text{2}}$ .

Formula used:

Work done $\left(W\right)$ by the force $\left(F\right)$ by moving a body through a distance $\left(\text{d}\right)$ is given by,

  $W=Fd$

  $\Rightarrow F=\frac{W}{d}$  $\left(3\right)$

Calculation:

Substituting the numerical values in equation $\left(3\right)$ ,

  $\begin{array}{c}F=\frac{-1.44\times {10}^5\text{ kg}{\text{.m}}^{\text{2}}{\text{/s}}^{\text{2}}}{\text{0}\text{.5 m}}\\ =-\text{2}\text{.88}\times {10}^5\text{ kg}{\text{.m/s}}^{\text{2}}\end{array}$

  $\left|F\right|=\text{2}\text{.88}\times {10}^5\text{ N}$

Conclusion:

The magnitude of the force that pushed in the front of the car by $\text{50}\text{.0 cm}$ is $\text{2}\text{.88}\times {10}^5\text{ N}$ .