A car starts from rest on a hill at the position shown. When the car reaches the

Question

A car starts from rest on a hill at the position shown. When the car reaches the bottom of the hill, it collides with a spring-loaded guardrail.

SOLUTION

(A) Find the maximum spring compression, assuming no energy losses due to friction.

Apply conservation of mechanical energy. Initially, there is only gravitational potential energy, and at maximum compression of the guardrail, there is only spring potential energy.

Solve for x.

x =

=

= 0.51 m(B) Calculate the maximum acceleration of the car by the spring, neglecting friction.

Apply Newton's second law to the car.

ma = ?kx      ?      a = ?

= ?

= ?

Substitute values.

(C) If the compression of the guardrail is only 0.30 m, find the change in the mechanical energy due to friction.

Use the work–energy theorem.

LEARN MORE

REMARKS The answer to part (b) is about 40 times greater than the acceleration of gravity, so we'd better be wearing our seat belts. Note that the solution didn't require calculation of the velocity of the car.

QUESTION True or False: In the absence of energy losses due to friction, doubling the height of the hill doubles the maximum acceleration delivered by the spring.

TrueFalse    

PRACTICE IT

Use the worked example above to help you solve this problem. A 15,000 N car starts from rest and rolls down a hill from a height of 10.0 m (see figure). It then moves across a level surface and collides with a light spring-loaded guardrail.

Explanation / Answer

A) when the car is at the highet of 10 m,it will have a gravitational potential energy that would change into potential energy of the spring, because there is no energy loss in any form.

So we have

Mgh=1/2 kx² where k=spring constant and x is compression in spring

Put values we have mg=F=13000 N

13000*10=1/2*1.0*10^6*x²

X² =13*10^4*2/10^6

X²=26*10—²

X²=0.26

X=0.51 m. This is the compression in the spring.

B)=According to newtons 3rd law, the force F exerted by car on spring =force Kx exerted by spring on car so we have

F=ma=kx

As we have mg=13000 and g =9.8

So m =F/g =13000/9.8 =1326 kg

Put values

1326*a =1*10^6 *0.51

a =384.61 m/s². This is the max acceleration of the car.

C) :if there is the compression of x'=0.30 m only,then energy in the spring is 1/2kx'² =1/2*10^6 *(0.30)² =

1/2*10^6 *9*10—²=4.5*10?

Initial energy of car=mgh=13000*10=13*10^4

So the energy difference between the two is the energy lost

Energy loss by friction =13*10^4- 4.5 10^4 =8.5*10^4 joule.

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

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