I have a question based off of the response to the question I posted below. Sorry, i wanted to make sure that i clearly understood,so My question about the third solution is: Is the horizontal force zero N? or is it 721.3N?
A dragster and driver together have mass 918.8 kg. The dragster,starting from rest,attains a speed of 25.8 m/s in .53s .
Find the average acceleration of the dragsters during this time interval. Answer in units of m/s^2.
What is the size of the average force on the dragster during this time interval?
Assume the driver has a mass of 73.6kg. What horizontal force does the seat exert on the driver?
Here is my work,but i don't know if i did it correctly:
Vf=Vi+a(delta t)
25.8=0+a(.53)
48.68=a
918.8(48.68)=44,726.5=F
73.6(48.68)=3,582.8N
physics - Henry, Sunday, July 3, 2011 at 6:14pm
a = 25.8 / 53 = 0.487m/s
F = 918.8*0.487 = 447.5N.
Fd = mg = 73.6kg * 9.8N/kg = 721.3N.
The angle between seat and hor = 0 deg.
Fh = 721.3sin(0) = 0 Newtons.
Fv = 721.3cos(0) = 721.3N = ver. force of driver.
Your method is correct.
In the given question, we need to calculate the horizontal force exerted on the driver by the seat. In this case, the size of the average force is equivalent to the vertical force exerted on the driver, which can be denoted as Fv.
To calculate Fv, we use the equation Fd = mg, where m is the mass of the driver (73.6 kg) and g is the acceleration due to gravity (approximately 9.8 m/s^2).
Fd = 73.6 kg * 9.8 m/s^2 = 721.3 N
So, the vertical force exerted on the driver is 721.3 N.
From the given information, we can determine that the angle between the seat and the horizontal direction is 0 degrees, so the horizontal force (Fh) exerted on the driver is zero Newtons.
Therefore, the answer is that the horizontal force exerted on the driver is zero N.