Physics for Scientists and Engineers with Modern Physics, Technology Update
Physics for Scientists and Engineers with Modern Physics, Technology Update
9th Edition
ISBN: 9781305401969
Author: SERWAY, Raymond A.; Jewett, John W.
Publisher: Cengage Learning
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Chapter 5, Problem 76AP

A 1.00-kg glider on a horizontal air track is pulled by a string at an angle θ. The taut string runs over a pulley and is attached to a hanging object of mass 0.500 kg as shown in Figure P5.40. (a) Show that the speed vx of the glider and the speed vy of the hanging object are related by vx = uvy, where u = z(z2h02)−1/2. (b) The glider is released from rest. Show that at that instant the acceleration ax of the glider and the acceleration ay of the hanging object are related by ax = uay. (c) Find the tension in the string at the instant the glider is released for h0 = 80.0 cm and θ = 30.0°.

Figure P5.40

Chapter 5, Problem 76AP, A 1.00-kg glider on a horizontal air track is pulled by a string at an angle . The taut string runs

(a)

Expert Solution
Check Mark
To determine

To show: The speed of the glider and the speed of the hanging object are related by vx=uvy where u=z(z2h02)(12) .

Answer to Problem 76AP

The speed of the glider and the speed of the hanging object are related by vx=uvy where u=z(z2h02)(12) .

Explanation of Solution

Given info: The mass of the glider is 1.00kg , the angle between the string and horizontal is θ , the mass of the hanging object is 0.500kg .

The free body diagram of the given case is as shown below.

Physics for Scientists and Engineers with Modern Physics, Technology Update, Chapter 5, Problem 76AP

Figure (1)

Form the above figure (1).

From the Pythagorean Theorem,

z2=x2+(h0)2

Here,

z is the length of string.

x is the distance of the glider on the ruler scale.

h0 is the string length that is holding the hanging object.

Rearrange the above equation for x .

x=(z2(h0)2)12

The speed of the glider is.

vx=dxdt

Substitute (z2(h0)2)12 for x in the above equation.

vx=ddt((z2(h0)2)12)=12(z2(h0)2)(12)2zdzdt . (1)

The term dzdt in the above expression is the rate of the string passing over the pulley so,

vy=dzdt

Here,

vy is the speed of the hanging object.

Substitute vy for dzdt in the equation (1).

vx=12(z2(h0)2)(12)2z(vy)=z(z2(h0)2)(12)(vy)

Substitute u for z(z2(h0)2)(12) in the above equation.

vx=u(vy) (2)

Conclusion:

Therefore, the speed of the glider and the speed of the hanging object are related by vx=uvy where u=z(z2h02)(12) .

(b)

Expert Solution
Check Mark
To determine

To show: The acceleration of the glider and the speed of the hanging object are related by. ax=uay .

Answer to Problem 76AP

The acceleration of the glider and the speed of the hanging object are related by. ax=uay .

Explanation of Solution

Given info: The mass of the glider is 1.00kg , the angle between the string and horizontal is θ , the mass of the hanging object is 0.500kg .

From equation (2), the relation of vx and vy is given as,

vx=u(vy)

The acceleration of the glider is,

ax=ddtvx

Substitute u(vy) for vx in the above equation.

ax=ddt[u(vy)]=uddt(vy)+vydudt

The initial velocity is zero.

Substitute 0 for u and ay for ddt(vy) in the above equation.

ax=uay

Here,

ay is the acceleration of the hanging object.

Conclusion:

Therefore, the acceleration of the glider and the speed of the hanging object are related by. ax=uay .

(c)

Expert Solution
Check Mark
To determine

The tension of the string.

Answer to Problem 76AP

The tension of the string is 3.56N .

Explanation of Solution

Given info: The mass of the glider is 1.00kg , the angle between the string and horizontal is 30.0° , the mass of the hanging object is 0.500kg .

From the free body diagram in figure (1) the net direction in x direction is,

z=h0sinθ

From part (a) the value of u is,

u=z(z2h02)(12)

Substitute h0sinθ for z in the above equation.

u=h0sinθ((h0sinθ)2h02)(12)

Substitute 30.0° for θ and 80.0cm for h0 in the above equation.

u=80.0cmsin(30.0°)((80.0cmsin(30.0°))2(80.0cm)2)(12)=115cm×1m100cm=1.15m

Thus, the value of u is 1.15m .

The net force in y direction is,

Tmg=may

Here,

T is the tension in the wire.

m is the mass of the hanging block.

Substitute 0.5kg for m and 9.8m/s2 for g .

T(0.5kg)(9.8m/s2)=(0.5 kg)ay

Rearrange the above equation for ay .

ay=(T0.5kg)+9.8m/s2 (3)

The net force in the x direction is,

Tcosθ=Max

Here,

M is the mass of the glider.

Form part (b) substitute uay for ax in the above equation.

Tcosθ=MuayT=Muaycosθ

From equation (3) substitute (T0.5kg)+9.8m/s2 for ay in the above equation,

T=Mucosθ((T0.5kg)+9.8m/s2)

Substitute 1.00kg for M , 0.500kg for m and 1.15m for u in the above equation.

T=(1.00kg)(1.15m)cos(30.0°)((T0.5kg)+9.8m/s2)3.18T=11.3NT=3.56N

Conclusion:

Therefore, the tension in the string is 3.56N

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Chapter 5 Solutions

Physics for Scientists and Engineers with Modern Physics, Technology Update

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