A 1310-kg racecar is driving around a circular track of radius 153 m. In the instant shown in the figure, the vehicle has a forward speed of 31.7 m/s and is slowing at a rate of 5.35 m/s2. What is the magnitude of the net force acting on the vehicle at this time and the force\'s direction relative to the velocity?
Well, this racecar seems to be going in circles and slowing down. It's like me trying to lose weight while running on a hamster wheel – quite a challenge!
To find the net force acting on the car, we need to consider the centripetal force and the force of friction. The centripetal force is responsible for keeping the car moving in a circle, and in this case, it acts towards the center of the circle.
The force of friction, on the other hand, acts in the opposite direction of the car's velocity. It's like trying to drive with the emergency brake on – not recommended!
To find the magnitude of the net force, we can use Newton's second law, which states that force equals mass times acceleration. In this case, the acceleration is the centripetal acceleration plus the deceleration due to slowing down:
a_net = a_circular + a_deceleration
The centripetal acceleration is given by a_circular = v^2 / r, where v is the car's velocity and r is the radius of the circular track.
Now, let's calculate the centripetal acceleration:
a_circular = (31.7 m/s)^2 / 153 m
And let's calculate the net acceleration:
a_net = (31.7 m/s)^2 / 153 m - 5.35 m/s^2
Finally, we can find the net force:
F_net = m * a_net
Plugging in the values, we get:
F_net = 1310 kg * [(31.7 m/s)^2 / 153 m - 5.35 m/s^2]
Calculating this gives us the magnitude of the net force acting on the vehicle. As for the direction, it depends on the orientation of the car's velocity and the center of the circular track.
Remember: don't try to drive in a circle while pulling the emergency brake. It won't end well!
To find the magnitude of the net force acting on the racecar, we can use Newton's second law of motion, which states that the net force acting on an object is equal to the mass of the object multiplied by its acceleration:
Net force = mass * acceleration
Given:
Mass of the racecar (m) = 1310 kg
Acceleration (a) = -5.35 m/s^2 (negative sign indicates deceleration)
We can calculate the net force as follows:
Net force = 1310 kg * (-5.35 m/s^2)
Net force = -7008.5 kg * m/s^2
Since we are asked to find the magnitude of the net force, we need to ignore the negative sign as magnitude is always positive:
Magnitude of the net force = 7008.5 N
Now let's determine the force's direction relative to the velocity. The force acts in the same direction as acceleration. In this case, the car is slowing down, so the net force is directed opposite to the velocity. Thus, the direction of the force is the exact opposite of the direction of velocity.
To find the magnitude of the net force acting on the vehicle and its direction relative to the velocity, we can start by considering the forces acting on the racecar.
The main force acting on the racecar is the net force, which can be calculated using Newton's second law: F = m * a, where F is the net force, m is the mass of the racecar, and a is its acceleration.
In this case, the racecar is experiencing centripetal acceleration as it moves in a circular path. The centripetal acceleration can be calculated using the equation: a = v^2 / r, where v is the velocity of the racecar and r is the radius of the circular track.
Given:
Mass of the racecar, m = 1310 kg
Radius of the circular track, r = 153 m
Forward speed of the racecar, v = 31.7 m/s
Acceleration of the racecar, a = -5.35 m/s^2 (negative sign implies deceleration)
To find the magnitude of the net force, we need to calculate the centripetal acceleration first:
a = v^2 / r
a = (31.7 m/s)^2 / 153 m
a = 10.63 m/s^2
Now, we can calculate the net force:
F = m * a
F = 1310 kg * 10.63 m/s^2
F = 13891.3 N
So, the magnitude of the net force acting on the vehicle at this time is 13891.3 N.
Now, let's determine the direction of the net force relative to the velocity. Since the racecar is slowing down, the net force must be pointing opposite to its velocity. Therefore, the direction of the net force is opposite to the direction of the forward velocity of the racecar.
In the given figure, the velocity vector is shown as an arrow pointing to the right. Therefore, the net force would be pointing to the left.
Normal (or centripetal) acceleration is
a(n) = v²/R = 31.7²/153=6.57 m/s²
Tangential acceleration
a(τ)= -5.35 m/s²
Acceleration
a=sqrt{ a(n)²+ (5.35)²} =8.47 m/s²
tanφ=5.35/6.57=0.814
φ=31.2º
α=90+φ=90 º +31.2 º =121.2 º