A car of 1000 kg with good tires on a dry road can decelerate (slow down) at a steady rate of about 5.0 m/s2 when braking. If a
1 answer:
(a) 4.0 s
The acceleration of the car is given by
where
v is the final velocity
u is the initial velocity
t is the time interval
For this car, we have
v = 0 (the final speed is zero since the car comes to a stop)
u = 20 m/s is the initial velocity
is the deceleration of the car
Solving the equation for t, we find the time needed to stop the car:
(b) 40 m
The stopping distance of the car can be calculated by using the equation
where
v = 0 is the final velocity
u = 20 m/s is the initial velocity
a = -5.0 m/s^2 is the acceleration of the car
d is the stopping distance
Solving the equation for d, we find
(c)
The deceleration is given by the problem, and its value is .
(d) 5000 N
The net force applied on the car is given by
where
m is the mass of the car
a is the magnitude of the acceleration
For this car, we have
m = 1000 kg is the mass
is the magnitude of the acceleration
Solving the formula, we find
(e)
The work done by the force applied by the car is
where
F is the force applied
d is the total distance covered
Here we have
F = 5000 N
d = 40 m (stopping distance)
So, the work done is
(f) The kinetic energy is converted into thermal energy
Explanation:
when the breaks are applied, the wheels stop rotating. The car slows down, as a result of the frictional forces between the brakes and the tires and between the tires and the road. Due to the presence of these frictional forces, the kinetic energy is converted into thermal energy/heat, until the kinetic energy of the car becomes zero (this occurs when the car comes to a stop, when v = 0).
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Answer:
a) 0 b) 29.9 N c) 21.7 N d) -17.4 N e) -13.0 N f) -17.4 N g) 16.9 N
h) 24.3 N θ = 44.2º
Explanation:
a) As the electric force is exerted along the line that joins the charges, due to any of the charges A or C has non-zero x-coordinates, the force has no x components either.
So, Fcax = 0
b) Similarly, as Fx = 0, the entire force is directed along the y-axis, and is going upward, due both charges repel each other.
Fyca = k*qa*qc / rac² = (9.10⁹ N*m²/C²*(2.79)*(1.07)*10⁻⁸ C²) / 9.00 m²
Fyca = 29. 9 N
c) In order to get the magnitude of the force exerted by B on C, we need to know first the distance between both charges:
rbc² = (3.00 m)² + (4.00m)² = 25.0 m²
⇒ Fbc = k*qb*qc / rbc² = (9.10⁹ N*m²/C²*(5.64)*(1.07)*10⁻⁸ C²) / 25.0 m²
⇒ Fbc = 21.7 N
d) In order to get the x component of Fbc, we need to get the projection of Fcb over the x axis, taking into account that the force on particle C is attractive, as follows:
Fbcₓ = Fbc * cos (-θ) where θ, is the angle that makes the line of action of the force, with the x-axis, so we can write:
cos θ = x/r = 4.00 / 5.00 m =
Fcbx = 21.7*(-0.8) = -17.4 N
e) The y component can be calculated in the same way, projecting the force over the y-axis, as follows:
Fcby = Fcb* sin (-θ) = 21.7* (-3.00/5.00) = -13.0 N
f) The sum of both x components gives :
Fcx = 0 + (-17.4 N) = -17.4 N
g) The sum of both y components gives :
Fcy = 29.9 N + (-13.0 N) = 16.9 N
h) The magnitude of the resultant electric force acting on C, can be found just applying Pythagorean Theorem, as follows:
Fc = √(Fcx)²+(Fcy)² = (17.4)² + (16.9)² = 24.3 N
The angle from the horizontal can be found as follows:
Ф = arc tg (16.9 / 17.4) = 44.2º