The acceleration of gravity is 9.8 m/s^2 . What is the magnitude of the total force on a(n) 72 kg driver by the dragster he operates as it accelerates horizontally along a straight line from rest to 61 m/s in 7.8 s? Answer in units of kN
F=m*a
It only asks about the force from the dragster, so ignore gravity. F = 72*a
a = (Vf-Vo)/t
a=61/7.8
F = (72)(71/7.8)
To find the magnitude of the total force on the driver, we can use Newton's second law of motion, which states that the force acting on an object is equal to the mass of the object multiplied by its acceleration.
First, let's calculate the acceleration of the dragster. We are given the initial velocity (0 m/s), the final velocity (61 m/s), and the time taken to reach that velocity (7.8 s). The formula to calculate acceleration is:
acceleration = (final velocity - initial velocity) / time taken
Substituting the given values:
acceleration = (61 m/s - 0 m/s) / 7.8 s
acceleration = 61 m/s / 7.8 s
Now, we can calculate the force using Newton's second law, where force = mass x acceleration. The mass of the driver is given as 72 kg:
force = mass x acceleration
force = 72 kg x acceleration
To convert the force to kilonewtons (kN), we divide by 1000 since 1 kN is equal to 1000 N:
force in kN = (72 kg x acceleration) / 1000
Now, we can substitute the value of acceleration into the equation to find the force in kilonewtons:
force in kN = (72 kg x (61 m/s) / 7.8 s ) / 1000
Calculating this equation will yield the magnitude of the total force exerted on the driver by the dragster as it accelerates horizontally along a straight line from rest to 61 m/s in 7.8 s.
Vinny I'm sorry to do this to someone who shares my name but bob is right ;3
total force? that is weight, plus horizontal force.
weight: 72*9.8 N
horizontal force= 72*(61/7.8) N
magnitude then= sqrt (weight^2+horizonal^2)