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

Expert's answer

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!

LATEST TUTORIALS
APPROVED BY CLIENTS