Respuesta :
1) 980,000 J; Â 0 J
2) 900,000 J; 400,000 J
3) 80,000 J; converted into thermal energy
4) -12,000 N
5) 180,000 W
6) 60,000 kg m/s; 40,000 kg m/s
7) 20 m/s
Explanation:
1)
The potential energy of an object is the energy possessed by the object due to its location in the gravitational field; it is given by
[tex]PE=mgh[/tex]
where
m is the mass
[tex]g=9.8 m/s^2[/tex] is the acceleration due to gravity
h is the height above the ground
For car 1, when it is at the top of the hill,
m = 2000 kg
h = 50 m
So, its potential energy is
[tex]PE=(2000)(9.8)(50)=980,000 J[/tex]
Car 2 instead was at the bottom of the hill, so h = 0, therefore its P.E. is zero:
PE = 0
2)
The kinetic energy of an object is the energy possessed by the object due to its motion. It is given by
[tex]KE=\frac{1}{2}mv^2[/tex]
where
m is the mass of the object
v is its speed
For car 1, at the bottom of the hill, we have
m = 2000 kg
v = 30 m/s
So its kinetic energy is:
[tex]KE=\frac{1}{2}(2000)(30)^2=900,000 J[/tex]
For car 2, we have
m = 2000 kg
v = 20 m/s
So its kinetic energy is
[tex]KE=\frac{1}{2}(2000)(20)^2=400,000 J[/tex]
3)
At the top of the hill, the total mechanical energy (sum of potential+kinetic energy) of car 1 was just equal to the initial potential energy:
[tex]E_i = PE=980,000 J[/tex]
At the bottom of the hill, the total mechanical energy is just equal to the final kinetic energy, so
[tex]E_f=KE=900,000 J[/tex]
Therefore, the energy lost is
[tex]\Delta E=E_i-E_f=980,000-900,000=80,000J[/tex]
The energy lost it has actually been converted into thermal energy, because of the presence of frictional forces that act against the motion of the car, slowing it down and converting part of its mechanical energy into kinetic energy.
4)
When car 1 applied force on its brakes, the work done on the car by the force is equal to its change in kinetic energy (work-energy theorem), so:
[tex]W=Fd=\frac{1}{2}mv^2-\frac{1}{2}mu^2[/tex]
where:
F is the force applied
d = 75 m is the distance covered while  stopping
v = 0 is the final velocity (the car comes to a stop)
u = 30 m/s is the initial velocity when the car starts slowing down
m = 2000 kg is the mass of the car
Solving for F, we find:
[tex]F=\frac{m}{2d}(v^2-u^2)=\frac{2000}{2(75)}(0^2-30^2)=-12,000 N[/tex]
where the negative sign indicates that the direction of the force is opposite to the direction of motion.
5)
The power used by car 1 to stop is given by
[tex]P=\frac{\Delta E}{t}[/tex]
where:
[tex]\Delta E[/tex] is the change in energy of the car  while it stops
t is the time interval required for the car to stop
In this problem, we have:
[tex]\Delta E=\frac{1}{2}mv^2-\frac{1}{2}mu^2=0-\frac{1}{2}(2000)(30)^2=900,000 J[/tex] is the magnitude of the kinetic energy lost by the car
t = 5 s is the time interval needed for the car to stop
Therefore, the power is
[tex]P=\frac{900,000}{5}=180,000 W[/tex]
6)
The momentum of an object is given by
[tex]p=mv[/tex]
where
m is the mass of the object
v is its velocity
In this problem, for car 1 before the accident, we have
m = 2000 kg
v = 30 m/s
So, its momentum was
[tex]p_1 = (2000)(30)=60,000 kg m/s[/tex]
For car 2, before the accident we have
m = 2000 kg
v = 20 m/s
So, its momentum was
[tex]p_2=(2000)(20)=40,000 kg m/s[/tex]
7)
Before the collision between the two cars:
- Car 1 has stopped, so its initial momentum is zero: [tex]p_1 =0[/tex]
- Car 2 is moving at 20 m/s, so its initial momentum is [tex]p_2=(2000)(20)=40,000 kg m/s[/tex]
After the collision:
- Car 2 has stopped, so its final momentum is [tex]p_2'=0[/tex]
- The final  velocity v of car 1 is unknown, so we can write its final momentum as [tex]p_1'=mv[/tex]
In a collision, the total momentum of the system is conserved, so we can write:
[tex]p_1+p_2=mv+p_2'[/tex]
And by solving for v, we find the final velocity of car 1:
[tex]v=\frac{p_2}{m}=\frac{40,000}{2000}=20 m/s[/tex]
