A 4.00 103-kg car has a speed of 9 m/s. The car then hits a tree. The tree doesn't move and the car comes to rest.
(a) Find the change in kinetic energy of the car.
. J
(b) Find the amount of work done in pushing in the front of the car.
J
(c) Find the size of the force that pushed the front of the car in 50 cm.
- i got -2.00e+05 for the first answer and -4.00e+05 for the third, but neither worked?
ΔKE =KE2 – KE1 = 0 - m•v²/2 =
= 4000• 9² /2 = - 162000 J.
a =v² /2•s =9² /2• 0.5= 81 m/s²
F =ma =4000•81 = 324 000 N
To find the answers to these questions, we need to apply the principles of work and energy.
(a) First, let's find the initial kinetic energy (KE) of the car using the formula:
KE = 0.5 * mass * (velocity)^2
Given:
Mass (m) = 4.00 x 10^3 kg
Velocity (v) = 9 m/s
Plugging in these values:
KE_initial = 0.5 * 4.00 x 10^3 kg * (9 m/s)^2
Simplifying:
KE_initial = 0.5 * 4.00 x 10^3 kg * 81 m^2/s^2
KE_initial = 16.2 x 10^3 kg m^2/s^2
KE_initial = 1.62 x 10^4 J
Now, since the car comes to rest after hitting the tree, the final kinetic energy is zero (KE_final = 0 J). Thus, the change in kinetic energy (ΔKE) is:
ΔKE = KE_final - KE_initial
ΔKE = 0 J - 1.62 x 10^4 J
ΔKE = -1.62 x 10^4 J
Therefore, the correct answer for the change in kinetic energy is -1.62 x 10^4 J.
(b) The work done in stopping the car is equal in magnitude but opposite in sign to the change in kinetic energy. Therefore, the work done is also -1.62 x 10^4 J.
However, in your response, you mentioned -4.00e+05 which doesn't seem to be related to this question.
(c) To find the force that pushed the front of the car a distance of 50 cm, we can use the work-energy theorem:
Work = Force * Distance
Given:
Work (W) = -1.62 x 10^4 J (from previous calculations)
Distance (d) = 50 cm = 0.5 m
Rearranging the formula:
Force = Work / Distance
Plugging in the values:
Force = (-1.62 x 10^4 J) / (0.5 m)
Force = -3.24 x 10^4 N
So, the size of the force that pushed the front of the car 50 cm is -3.24 x 10^4 N.
Please note that the negative sign indicates that the force is in the opposite direction of the car's motion.