PHYSICS FOR SCI.AND ENGR W/WEBASSIGN
PHYSICS FOR SCI.AND ENGR W/WEBASSIGN
10th Edition
ISBN: 9781337888462
Author: SERWAY
Publisher: CENGAGE L
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Chapter 5, Problem 40AP

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 40AP, 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

The relation between the speed of the glider and the speed of the hanging object.

Answer to Problem 40AP

The relation between the speed of the glider and the speed of the hanging object is vx=uvy where u=z(z2h02)(12).

Explanation of Solution

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 SCI.AND ENGR W/WEBASSIGN, Chapter 5, Problem 40AP

Figure (1)

Form the above figure (1).

Write the expression for the length of the string using Pythagorean Theorem,

    z2=x2+(h0)2

Here, z is the length of string, x is the distance of the glider on the ruler scale and h0 is the string length that is holding the hanging object.

Rearrange the above equation for x.

    x=(z2(h0)2)12

Write the expression for the speed of the glider

    vx=dxdt

Here, vx is the speed of the glider.

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

    vx=ddt((z2(h0)2)12)=12(z2(h0)2)(12)2zdzdt                                                                       (I)

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

Write the expression for the speed of the hanging object.

    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)                                                                                                  (II)

Conclusion:

Therefore, the relation between the speed of the glider and the speed of the hanging object is vx=uvy where u=z(z2h02)(12).

(b)

Expert Solution
Check Mark
To determine

The relation between the acceleration of the glider and the speed of the hanging object.

Answer to Problem 40AP

The relation between the acceleration of the glider and the speed of the hanging object is ax=uay.

Explanation of Solution

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

    vx=u(vy)

Write the expression for the acceleration of the glider

    ax=ddtvx

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

    ax=ddt[u(vy)]=uddt(vy)+vydudt

The initial velocity of the hanging object is zero.

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

    ax=uay

Here, ay is the acceleration of the hanging object.

Conclusion:

Therefore, the relation between the acceleration of the glider and the speed of the hanging object is ax=uay.

(c)

Expert Solution
Check Mark
To determine

The tension of the string.

Answer to Problem 40AP

The tension of the string is 3.56N.

Explanation of Solution

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

    z=h0sinθ

From part (a) the value of u

    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

    Tmg=may

Here, T  is the tension in the wire and 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                                                                        (III)

The net force in the x direction

    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)

Conclusion:

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)3.655T=13.013T=3.56N

Therefore, the tension in the string is 3.56N.

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Chapter 5 Solutions

PHYSICS FOR SCI.AND ENGR W/WEBASSIGN

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