Question

a. What is the kinetic energy of a $1500kg$car traveling at a speed of $30m/s(\approx 65mph)$?

b. From what height would the car have to be dropped to have this same amount of kinetic energy just before impact?

Short answer

(a) As a result, the car's kinetic energy is $6.75\times {10}^5\mathrm{J}$.

(b) The car's required height, rounded up to two major digits, is$46m.$

Step-by-step solution

Introduction 

The energy that an item or particle has as a result of its movement is known as kinetic energy. When a net force is applied to an item, the object accelerates and accumulates kinetic energy as a result.

Explanation (a)

Kinetic energy is the energy possessed by a body as a result of its motion.

$K=\frac{1}{2}m{v}^2$

Here, $m$is mass of the body and $v$is velocity of the body.

Substitute $1500\mathrm{kg}$for $m$and $30\mathrm{m}/\mathrm{s}$for $v$in the equation $K=\frac{1}{2}m{v}^2$to find the kinetic energy of the car.

$K_{\mathrm{f}}=\frac{1}{2}(1500\mathrm{kg})(30\mathrm{m}/\mathrm{s}{)}^2$

$=6.75\times {10}^5\mathrm{J}$

As a result, the car's kinetic energy is$6.75\times {10}^5\mathrm{J}$.

Explanation (b)

Equate the expressions for total energy at height $y_{\mathrm{i}}$and the kinetic energy at the height $y_{\mathrm{f}}$on the ground.

$K_{\mathrm{f}}+mg{y}_{\mathrm{f}}=K_{\mathrm{i}}+mg{y}_{\mathrm{i}}$

Here, $K_{\mathrm{f}}$is final kinetic energy at height $y_{\mathrm{f}}$and $K_{\mathrm{i}}$is initial kinetic energy on the ground.

Rewrite the equation for $y_{\mathrm{i}}$.

$K_{\mathrm{f}}-K_{\mathrm{i}}+mg{y}_{\mathrm{f}}=mg{y}_{\mathrm{i}}$

$y_{\mathrm{i}}=\frac{\left(K_{\mathrm{f}}-K_{\mathrm{i}}\right)+mg{y}_{\mathrm{f}}}{mg}$

Substitute $6.75\times {10}^6\mathrm{J}$for $K_{\mathrm{f}},0\mathrm{J}$for $K_{\mathrm{i}},0\mathrm{m}$for $y_{\mathrm{f}},1500\mathrm{kg}$for $m$, and $9.8\mathrm{m}/{\mathrm{s}}^2$for $g$in the above equation.

$y_{\mathrm{i}}=\frac{\left(6.75\times {10}^6\mathrm{J}-0\mathrm{J}\right)+(1500\mathrm{kg})\left(9.8\mathrm{m}/{\mathrm{s}}^2\right)(0\mathrm{m})}{(1500\mathrm{kg})\left(9.8\mathrm{m}/{\mathrm{s}}^2\right)}$

$=45.92\mathrm{m}$

The car's required height, rounded up to two major digits, is$46\mathrm{m}$.

Sub part of b

Simplify the expression for height in part (b) in further.

$y_{\mathrm{i}}=\frac{\left(K_{\mathrm{f}}-K_{\mathrm{i}}\right)+mg{y}_{\mathrm{f}}}{mg}$

Substitute $\frac{1}{2}m{v}_{\mathrm{f}}^2$for $K_{\mathrm{f}}$and $\frac{1}{2}m{v}_{\mathrm{i}}^2$for $K_{\mathrm{i}}$.

$y_{\mathrm{i}}=\frac{\left(\frac{1}{2}m{v}_{\mathrm{f}}^2-\frac{1}{2}m{v}_{\mathrm{i}}^2\right)+mg{y}_{\mathrm{f}}}{mg}$

$=\frac{m\left(\frac{1}{2}{v}_{\mathrm{f}}^2-\frac{1}{2}{v}_{\mathrm{i}}^2\right)+g{y}_{\mathrm{f}}}{mg}$

$=\frac{\left(\frac{1}{2}{v}_{\mathrm{f}}^2-\frac{1}{2}{v}_{\mathrm{i}}^2\right)+g{y}_{\mathrm{f}}}{g}$

It is obvious from the preceding formulation that the required height$y_{\mathrm{i}}$ is independent of the car's mass. As a result, we may deduce that the answer to section (b) is independent of the car's mass.