COLLEGE PHYSICS
COLLEGE PHYSICS
2nd Edition
ISBN: 9781464196393
Author: Freedman
Publisher: MAC HIGHER
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Chapter 6, Problem 97QAP
To determine

(a)

The distance up to which the second ramp object slide before coming to a momentary stop.

Expert Solution
Check Mark

Answer to Problem 97QAP

The distance up to which the second ramp object slide before coming to a momentary stop is 19.94m.

Explanation of Solution

Given:

The given figure is,

COLLEGE PHYSICS, Chapter 6, Problem 97QAP , additional homework tip  1

Figure 1

The values are:

  H1=12.0mH2=7.00mθ1=60.0°θ2=37.0°

Concept Used:

Interpreting kinetic energy.

Calculation:

Let us consider the kinetic energy at the starting point:

  Ki=12mvi2=0Ugrav,i=mgyi=mgH1

At the point where the second ramp ends kinetic energy is:

  Kf=12mvf2Ugrav,f=mgyf=mgdsinθ2

According to the law of conservation of energy:

  Ui+Ki=Uf+Kf0+mgH1=mgdsinθ2+0d=H1sinθ2

On replacing the values, we get

  H1=12.0mθ2=37.0°d=12.0sin37.0d=19.94m

Conclusion:

Thus, the distance up to which the second ramp object slide before coming to a momentary stop is 19.94m.

To determine

(b)

The speed of the object at the height of H2=7.00m.

Expert Solution
Check Mark

Answer to Problem 97QAP

The speed of the object at the height of H2=7.00m is 9.90m/sec.

Explanation of Solution

Given:

The given figure is,

COLLEGE PHYSICS, Chapter 6, Problem 97QAP , additional homework tip  2

Figure 1

The values are:

  H1=12.0mH2=7.00mθ1=60.0°θ2=37.0°

Concept Used:

Interpreting potential energy.

Calculation:

Let us consider the kinetic energy at the starting point:

  Ki=12mvi2=0Ugrav,i=mgyi=mgH1

At the point where the second ramp ends kinetic energy is:

  Kf=12mvf2Ugrav,f=mgyf=mgH2

According to the law of conservation of energy:

  Ui+Ki=Uf+Kf0+mgH1=mgH2+12mvf2vf=2g(H1H2)

On replacing the values, we get

  H1=12.0mH2=7.00mg=9.8m/sec2vf=2×9.8(12.07.00)vf=9.90m/sec

Conclusion:

The speed of the object at the height of H2=7.00m is 9.90m/sec.

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Chapter 6 Solutions

COLLEGE PHYSICS

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