Question

A car's bumper is designed to withstand a 4.32 km/h (1.2 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 0.150 m while bringing a 820 kg car to rest from an initial speed of 1.2 m/s. (Enter a number.)

Answer 1

`F=3936\ N`

Step-by-step explanation:

Given:

• initial speed of the car,
  `u=1.2\ m.s^(-1)`

• mass of the car,
  `m=820\ kg`

• final speed of the car,
  `v=0\ m.s^(-1)`

• distance collapsed due to collision,
  `s=0.15\ m`

Now by the law of energy conservation, the work done is the energy compensated in bringing the car to rest.

`W=\Delta KE`

`F.s=(1)/(2) m.u^2-(1)/(2) m.v^2`

where:

`W=` work done by the force F

`\Delta KE=` change in kinetic energy

`F* 0.15=(1)/(2)* 820* 1.2^2-0`

`F=3936\ N`

Answer 2

3936 N

Step-by-step explanation:

m = Mass of car = 820 kg

u = Initial velocity = 1.2 m/s

v = Final velocity = 0

d = Distance = 0.15 m

The work done is given by the change in the kinetic energy of the system

`W=(1)/(2)m(v^2-u^2)\\\Rightarrow Fd=(1)/(2)m(v^2-u^2)\\\Rightarrow F=((1)/(2)m(v^2-u^2))/(d)\\\Rightarrow F=((1)/(2)820(0^2-1.2^2))/(0.15)\\\Rightarrow F=-3936\ N`

The magnitude of force is 3936 N