The acceleration of gravity is 9.8 m/s2 . What is the magnitude of the total force on a(n) 75 kg driver by the dragster he operates as it accelerates horizontally along a straight
line from rest to 66 m/s in 4.5 s? Answer in units of kN
a = (V-Vo)/t = (66-0)/4.5 = 14.67 m/s^2.
F = M*a
1100.25
To find the magnitude of the total force on the driver, we need to consider the acceleration and the mass of the driver.
Given:
Acceleration of gravity (g) = 9.8 m/s^2
Mass of the driver (m) = 75 kg
Final velocity (v) = 66 m/s
Time taken (t) = 4.5 s
First, let's calculate the acceleration (a) using the equation:
a = (v - u) / t
Where:
v = final velocity = 66 m/s
u = initial velocity = 0 m/s (since the car starts from rest)
t = time taken = 4.5 s
a = (66 - 0) / 4.5 = 14.67 m/s^2
Now, we can calculate the net force (F) acting on the driver using Newton's second law:
F = m * a
Where:
m = mass of the driver = 75 kg
a = acceleration = 14.67 m/s^2
F = 75 kg * 14.67 m/s^2 = 1100.25 N
Since the question asks for the magnitude of the total force in kilonewtons (kN), we can convert the force from Newtons to kilonewtons by dividing it by 1000:
F_kN = F / 1000 = 1100.25 / 1000 = 1.10025 kN
Therefore, the magnitude of the total force on the 75 kg driver is approximately 1.10025 kN.