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4vir4ik [10]
3 years ago
13

A 3.00-kg box that is several hundred meters above the earth’s surface is suspended from the end of a short vertical rope of neg

ligible mass. A time-dependent upward force is applied to the upper end of the rope and results in a tension in the rope of T(t) = (38.0 N/s)t. The box is at rest at t = 0. The only forces on the box are the tension in the rope and gravity. A) What is the velocity of the box at t = 1.00 s? B) What is the maximum distance that the box descends below its initial position? C) At what value of t does the box return to its initial position? D) What is the velocity of the box at t = 3.00 s?
Physics
1 answer:
marshall27 [118]3 years ago
6 0

Answer:

A) 3.48m/s

B) 3.92m

C) 2.32m

D 23.33m/s

Explanation:

ma(t)=mg-At

a(t)= g - (At/m)

V(t)= integrala(T)dT = gt- (At^2/2m)

Initial x coordinator of the box is zero

X(t)= integralV(t)dt= 1/2gt^2-(At^3/6m)

a) V =( 9.81×1) -(38×1^2/2×3)

V= 9.81-6.33= 3.48m/s

b)-AT^2/2m + gT= 0

T=2mg/A= (2×3×9.81)/38

T= 1.549m

X(T)= (1/2×9.81×1.549^3)- (38×1.549^3/6×3)

X(T)= 11.768- (141.23/18) = 11.768 - 7.85= 3.92m

C) 1/2gT''^2 - AT''^3/6m =0

The only non trivial solution is T''= 3mg/A

T=(3×3×9.81)/38 = 2.32m

D) V = 9.81×3) - (38×3^2/6)

V= 29 - 5.667= 23.33m/s

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E_1=\frac{GM_em}{R_e}-\frac{GM_em}{R_e+h}

Now Energy required to orbit around the earth

E_2=\frac{1}{2}mv_{orbit}^2=\frac{GM_2m}{2(R_e+h)}

\Delta E=E_1-E_2

\Delta E=\frac{GM_em}{R_e}-\frac{GM_em}{R_e+h}-\frac{GM_2m}{2(R_e+h)}

E_1=E_2  (given)

\frac{GM_em}{R_e}-\frac{GM_em}{R_e+h}-\frac{GM_2m}{2(R_e+h)}=0

\frac{1}{R_e}-\frac{3}{2(R_e+h)}=0

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thus energy to lift the satellite is more than orbiting around earth

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To push a 26.0 kg crate up a frictionless incline, angled at 25.0° to the horizontal, a worker exerts a force of 209 N parallel
alukav5142 [94]

Answer:

(a) W = +397.1 J

(b) W = -204.6 J

(c) W = 0

(d) W= + 192.5 J

Explanation:

Work (W) is defined as the product of force (F) by the distance (d)the body travels due to this force. :

W= F*d Formula ( 1)

The forces that perform work on an object must be parallel to its displacement.

The forces perpendicular to the displacement of an object do not perform work on it.

The work is positive (W+) if the force has the same direction of movement of the object.  

The work is negative (W-) if the force has the opposite direction of the movement of the object.

Problem development

(a) Work performed by the worker's applied force on the box .

W= 209 N * 1.9 m = +397.1 J

(b) Work performed by the gravitational force on the crate

We calculate the weight component parallel to the displacement of the box:

We define the x-axis in the direction of the inclined plane ,25.0° to the horizontal.

We define the y-axis and in the direction of the plane perpendicular to the inclined plane.

W= m*g=26*9.8= 254.8N : total box weight

Wx= W*sen25.0°= 254.8*sen25.0°= 107.68 N

W = -Wx *d =107.68 N *1.9 m= -204.6 J

(c) Work performed by normal force (N) exerted by the incline on the crate

The force N is perpendicular to the displacement, then:

W=0

(d) Total work done on the crate

W = 397.1 J -204.6 J

W = 192.5 J

4 0
2 years ago
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