A 1500kg car traveling at a steady speed of 20 m/s initially due northwest round a corner so that after 10s, it is traveling due northeast. What is the magnitude and direction of the net force that must be acting on the car at the instant it is traveling due north?
To find the magnitude and direction of the net force acting on the car when it is traveling due north, we first need to understand the changes in velocity and direction.
Given:
Mass of the car (m) = 1500 kg
Initial speed (v1) = 20 m/s
Time (t) = 10 s
At the beginning, the car is traveling due northwest, which means the velocity vector is pointing in that direction. After 10 seconds, the car is traveling due northeast, implying a change in direction.
To determine the change in velocity, we can use the formula:
Δv = v2 - v1
where Δv is the change in velocity, v2 is the final velocity, and v1 is the initial velocity.
Since the car is traveling due north at the end, we know the final velocity is also pointing north. Therefore, the change in velocity is:
Δv = v2 - v1 = north - northwest = north + west
Now, we can calculate the change in velocity:
Δv = √[(north)^2 + (west)^2]
Using Pythagoras' theorem, we can calculate the magnitude of the change in velocity. The direction can be found using trigonometry:
Magnitude of Δv = √[0^2 + 20^2] = 20 m/s
To find the direction, we can use the trigonometric relationships in a right-angled triangle involving the north and west components of Δv.
Sin(θ) = opposite/hypotenuse = west/20
Cos(θ) = adjacent/hypotenuse = north/20
From these equations, we can solve for the west and north components of Δv:
west = 20 * sin(θ)
north = 20 * cos(θ)
Since θ is the angle between the initial velocity vector and the north direction, we calculate it as follows:
θ = tan^(-1)(west/north)
Next, we need to determine the net force acting on the car when it is traveling due north. The net force is responsible for changing the car's direction and velocity.
The net force acting on an object can be calculated using Newton's second law of motion:
Net force (F_net) = mass (m) * acceleration (a)
Since the car is traveling at a steady speed, the net force acting on the car is perpendicular to the velocity vector.
Now, we know that acceleration (a) is the rate at which the velocity changes with respect to time:
a = Δv/t
Plugging in the values, we find:
a = 20 m/s / 10 s = 2 m/s^2
Finally, we can determine the net force acting on the car at the instant it is traveling due north:
F_net = m * a = 1500 kg * 2 m/s^2 = 3000 N
Therefore, the magnitude of the net force acting on the car is 3000 N. However, we need to give the direction as well.
Since the net force is perpendicular to the velocity vector, it must be pointing in the east direction. Thus, the direction of the net force is due east.