A truck of mass 2 tonnes moving at a speed of 25 m/s is brought to rest suddenly in 5 seconds. Calculate the power exerted by the brakes

power=energy/time=(1/2 mv^2)/5

change tonnes to kg

work done = loss of ke = (1/2)m v^2

= 1000*625 Joules
power = work/second
= 1000 * 625 / 5 Watts

A man claims that he can hold onto a 9.0-kg child in a head-on collision as long as he has his seat belt on. Consider this man in a collision in which he is in one of two identical cars each traveling toward the other at 49.0 mi/h relative to the ground. The car in which he rides is brought to rest in 0.05 s.

(a) Find the magnitude of the average force needed to hold onto the child.

To calculate the power exerted by the brakes, we first need to determine the force exerted by the brakes to bring the truck to a stop. We can use Newton's second law of motion:

Force (F) = mass (m) × acceleration (a)

The acceleration (a) can be calculated using the equation:

a = Δv / Δt

Where Δv is the change in velocity and Δt is the time taken for that change.

Given that the truck is brought to rest suddenly, its final velocity (vf) is 0 m/s, and the initial velocity (vi) is 25 m/s. The time taken (Δt) to come to rest is 5 seconds.

Δv = vf - vi = 0 - 25 = -25 m/s

Using the equation for acceleration, we can calculate:

a = Δv / Δt
a = (-25 m/s) / (5 s)
a = -5 m/s^2

Now that we have the acceleration, we can calculate the force exerted by the brakes:

F = m × a
F = 2000 kg × (-5 m/s^2)
F = -10000 N

Since the force exerted by the brakes is negative (opposite to the direction of motion), we can take the absolute value:

|F| = 10000 N

Finally, to calculate the power exerted by the brakes, we use the formula:

Power (P) = Force (F) × velocity (v)

Given that velocity (v) is 25 m/s:

P = |F| × v
P = 10000 N × 25 m/s
P = 250000 W

Therefore, the power exerted by the brakes is 250,000 watts.