water leaves a hose at a rate of 1.5kg/s with a speed of 20m/s and is aimed at the side of a car, which stops it (splashing back is ignored). what is the force exerted by the water on the car?
To find the force exerted by the water on the car, we can use Newton's second law of motion, which states that the force acting on an object is equal to its mass multiplied by its acceleration.
In this case, the water leaving the hose is considered as a jet of water. The mass flow rate of the water is given as 1.5 kg/s, which means that 1.5 kilograms of water is expelled from the hose every second.
Now, when the water hits the car and stops, its final velocity becomes zero. We can assume that the water's acceleration is constant, as the change in velocity occurs over a very short time interval. Hence, we can calculate the acceleration of the water by using the following equation:
Final velocity (v_final) = Initial velocity (v_initial) + (acceleration (a) × time (t))
Since v_final = 0 (as the water stops) and v_initial = 20 m/s (given in the problem), we can rewrite the equation as:
0 = 20 m/s + (a × t)
We also know that the mass (m) of the water expelled in time t is 1.5 kg. Since force (F) is equal to mass multiplied by acceleration, we can rearrange the equation above to solve for acceleration:
a = -20 m/s / t
Now, substituting a = -20 m/s / t into the formula F = m × a, we can find the force exerted by the water on the car:
F = (1.5 kg) × (-20 m/s / t)
Therefore, the force exerted by the water on the car depends on the time interval (t) over which the water stops.