Question #121442
. A 9,100 kg tractor is moving (horizontally) at 5.0 km h−1 when the 80 kg driver has
to brake suddenly to come to a stop before hitting a tree stump. If the tractor and
driver come to rest over a horizontal distance of 2.0 m, and the acceleration of the
tractor and driver is constant, what is the magnitude of the net force on the driver
1
Expert's answer
2020-06-10T18:22:20-0400

Initial velocity is v0=5.0km/h=1.389m/sv_0=5.0 km/h= 1.389 m/s

Final velocity is v=0v=0

Distance traveled is d=2.0md=2.0 m

The acceleration is

a=v2v022d=(0)2(1.389)22(2.0)=0.482m/s2a=\frac{v^2-v_0^2}{2d}=\frac{(0)^2-(1.389)^2}{2(2.0)}=-0.482 m/s^2

the magnitude of the net force on the driver is

F=ma=(80kg)(0.482m/s2)=38.58N=38.6NF=ma=(80 kg)(0.482 m/s^2)=38.58 N=38.6 N


Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!
LATEST TUTORIALS
APPROVED BY CLIENTS