QUESTION 1 A 120 kg stone slides down a snowcovered hill the

QUESTION #1

A 12.0 kg stone slides down a snow-covered hill (the figure (Figure 1) ), leaving point A with a speed of 10.0 m/s . There is no friction on the hill between points A and B, but there is friction on the level ground at the bottom of the hill, between B and the wall. After entering the rough horizontal region, the stone travels 100 m and then runs into a very long, light spring with force constant 2.10 N/m . The coefficients of kinetic and static friction between the stone and the horizontal ground are 0.20 and 0.80, respectively.

Part A

What is the speed of the stone when it reaches point B?

23.15

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Part B

How far will the stone compress the spring?

x=

QUESTION #2

A truck with mass m has a brake failure while going down an icy mountain road of constant downward slope angle (Figure 1) . Initially the truck is moving downhill at speed v0. After careening downhill a distance L with negligible friction, the truck driver steers the runaway vehicle onto a runaway truck ramp of constant upward slope angle . The truck ramp has a soft sand surface for which the coefficient of rolling friction is r.

Part A
What is the distance that the truck moves up the ramp before coming to a halt? Solve using energy methods.
Express your answer in terms of m, , v0, L, g, and r.

QUESTION #3
A basket of negligible weight hangs from a vertical spring scale of force constant 1500 N/m .

Part A
If you suddenly put a 3.00 kg adobe brick in the basket, find the maximum distance that the spring will stretch.

lmax =
  cm  

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Part B
If, instead, you release the brick from 1.00 m above the basket, by how much will the spring stretch at its maximum elongation?

l =
  cm

Solution

Q1:

use the law of conservation of energy to find the speed.

     mgy + 0.5mv0^2 =0.5mv^2

   v= sqrt[v0^2 + 2gy] = sqrt[10*10 + 2*9.8*20] = 22.18 m/s

use the law of conservation of energy to find the compressed length.

               mgy + 0.5mv0^2 = 0.5kx^2

then the compressed length is,

      x = sqrt{[mv0^2 + 2mgy]/k} = 53.02 m