a car of mass 1000 kg moving at a speed of 20 m/s is brought to rest through a distance of 100 m by applying the brakes.a)calculate the accleration of the car.b) calculate the force by the brakes assuming that is constant and there is no other resistance.c)calculate the work done by the brakes.
a. V^2 = Vo^2 + 2a*d.
0 = 20^2 + 2a*100, a = -2 m/s^2.
b. F = M*a = 1000*(-2) = -2,000 N.
c. W = F*d = -2000 * 100 = -200,000 Joules.
qwd
To solve this problem, we can use the following equations:
a) Acceleration (a) can be calculated using the equation:
a = (v^2 - u^2) / (2s)
where v is the final velocity (0 m/s in this case),
u is the initial velocity (20 m/s in this case), and
s is the distance covered (100 m in this case).
b) Force (F) can be calculated using Newton's second law of motion:
F = m * a
where m is the mass of the car (1000 kg in this case) and
a is the calculated acceleration.
c) Work done (W) can be calculated using the equation:
W = F * s
where F is the calculated force and
s is the distance covered.
Now, let's calculate the answers to each part of the question:
a) Calculation of acceleration:
a = (0^2 - 20^2) / (2 * 100)
= (-400) / 200
= -2 m/s^2
Therefore, the acceleration of the car is -2 m/s^2.
b) Calculation of force:
F = m * a
= 1000 kg * (-2 m/s^2)
= -2000 N
Therefore, the force by the brakes is -2000 N. The negative sign indicates that the force is in the opposite direction to the motion of the car.
c) Calculation of work done:
W = F * s
= -2000 N * 100 m
= -200,000 J
Therefore, the work done by the brakes is -200,000 J. Again, the negative sign indicates that work is done against the motion of the car.
Note: The negative signs in the answers indicate that the direction of acceleration, force, and work is opposite to the initial motion of the car.