Question

A car's bumper is designed to withstand a 7.20 km/h (2.0 m/s) collision with an immovable object without damage to the body of the car. The bumper cushions the shock by absorbing the force over a distance. Calculate the magnitude of the average force (in N) on a bumper that collapses 0.195 m while bringing a 830 kg car to rest from an initial speed of 2.0 m/s.

Answer 1

Answer:

8512 N

Explanation:

From the work energy theorem we know that: The net work done on a particle equals the change in the particles kinetic energy:

$W=\Delta K=K_{f}-K_{i} \\ \\qquad \begin{array}{r} W=F \cdot d, \Delta K=\frac{1}{2} m v_{f}^{2}-\frac{1}{2} m v_{i}^{2} \\ F \cdot d=\frac{1}{2} m v_{f}^{2}-\frac{1}{2} m v_{i}^{2}$

Where:

-W is the work done by the force.

- F is the force actin on the.

- d is the distance travelled.

- m is the mass of the car.

- $v_{f}, v_{i}$ are the final and the initial velocity of the car

$K_{f}, K_{i}$ are the final and the kinetic energy of the car.

Givens: $m=830 \mathrm{~kg}, v_{i}=2 \mathrm{~m} / \mathrm{s}, v_{f}=0 \mathrm{~km} / \mathrm{h}, d=0.195 \mathrm{~m}$

Plugging known information to get:

$F \cdot d &=\frac{1}{2} m v_{f}^{2}-\frac{1}{2} m v_{i}^{2} \\ F &=\frac{\frac{1}{2} m v_{f}^{2}-\frac{1}{2} m v_{i}^{2}}{d} \\ &=\frac{0-\frac{1}{2} \times 830 \times 2^{2}}{0.195} \\ &=8.512 \times 10^{3} \\ F &=8.512 \times 10^{3} \mathrm{~N}$