An automobile of mass 2000 kg is driven into a brick wall in a safety test. The bumper behaves like a Hooke’s-law spring. It has an effective spring constant of 5 × 106 N/m, and is observed to compress a distance of 8.7 cm as the car is brought to rest. What was the initial speed of the automobile?
Answer in units of m/s
i used PE=(.5)(k)(d^2)
so
PE=.(.5)(5x10^6)(.087^2) =18922.5
then from there im not sure what do to, did i do that part and where do i go from there?
thanks
Great start! You correctly used the formula for potential energy (PE) of a spring, which is given by PE = (0.5)(k)(d^2), where k is the spring constant and d is the displacement.
In this case, the potential energy represents the initial kinetic energy (KE) of the automobile. Since the car comes to a stop, all its initial kinetic energy is stored as potential energy in the compressed spring.
Therefore, you have correctly calculated the potential energy as PE = 18922.5 J.
To find the initial speed of the automobile, you can equate the initial kinetic energy to the potential energy.
KE = PE
(0.5)(m)(v^2) = PE
Here, m is the mass of the automobile and v is its initial velocity.
Rearranging the equation to solve for v, we have:
v^2 = (2)(PE) / m
v = √[(2)(PE) / m]
Substituting the values, the equation becomes:
v = √[(2)(18922.5) / 2000]
Calculating this expression will give you the initial speed of the automobile.