Asked by phirum
A 1100 kg car coasts on a horizontal road with a speed of 19 m/s. After crossing a an unpaved, sandy stretch of road 32 m long, its speed decreases to 12 m/s. (a) Was the net force done on the car positive, negative, or zero? Explain (b) Find the magnitude of the average net force on the car in the sandy section.
Answers
Answered by
Muhlis Olcay
Equations:
Change in kinetic energy = 1/2m(Vf^2-Vi^2) = Net (total) Work done on the object = F*d
m = 1100 kg
Vi = 19 m/s
Vf = 12 m/s
d = 32m
Calculations:
1/2(1100)(12^2-19^2) = F(net)*32
=119350/32
b) =3730 N
a) ΔEK is negative, meaning that W < 0 meaning that theta > 90 degrees (cos180 for this question)
Change in kinetic energy = 1/2m(Vf^2-Vi^2) = Net (total) Work done on the object = F*d
m = 1100 kg
Vi = 19 m/s
Vf = 12 m/s
d = 32m
Calculations:
1/2(1100)(12^2-19^2) = F(net)*32
=119350/32
b) =3730 N
a) ΔEK is negative, meaning that W < 0 meaning that theta > 90 degrees (cos180 for this question)