4. A truck of mass m = 2500 kg is traveling downhill on a road with slope l = 3.

Question

4. A truck of mass m = 2500 kg is traveling downhill on a road with slope l = 3.5°. The driver slams on the brakes and skids to a stop, leaving skid marks 30 m long. The coefficient of friction between the truck's wheels and the road is u = 0.4. (a) (2 points) Using concepts of work and energy, find an expression for how fast the truck was going just before hitting the brakes. Was the truck exceeding the speed limit of 25 miles per hour? (b) (2 points) All else being equal, how far would the truck have skidded if the slope of the road were 0 = 5°? Is there a slope angle beyond which the truck would not have been able to stop at all? What is this angle? (c) (2 points) How do your answers in (a) and (b) change, if the truck had a mass 4000 kg instead of 2500 kg? Briefly discuss.

Explanation / Answer

mass, m=2500 kg

theta=3.5 degrees

distance, d=30m

friction coefficient, uk=0.4

a)

by using law of conservation enegry,

W=P.E+K.E

fK*d=m*g*h+1/2*m*v^2

uk*m*g*cos(theta)*d=m*g*d*sin(theta)+1/2*m*v^2

uk*g*cos(theta)*d = g*d*sin(theta)+1/2*v^2 ----(1)

0.4*9.8*cos(3.5)*30=9.8*30*sin(3.5)+1/2*v^2

===> v=14.1 m/sec or 31.54 miles/hr

speed of the truck just before hitting the breakes, v=14.1 m/sec

and truck exceeding the speed limit 25 miles per hour

b)

if theta=5 degrees,

from equation (1),

uk*g*cos(theta)*d = g*d*sin(theta)+1/2*v^2

0.4*9.8*cos(5)*d=9.8*d*sin(5)+1/2*14.1^2

==> d=32.58 m/sec

ditance travelled, d=32.58 m/sec

and

if

m*g*sin(theta')=fk then, the truck may not be stop,

m*g*sin(theta')=uk*m*g*cos(theta')

==> tan(theta')=uk

tan(theta')=0.4

===> theta'=21.8 degrees

c)

here,

speed v and distance d does not depend on the mass

hence, V and d will remain same

Related Questions

Navigate

• Browse (All)
• Browse 4
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.