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nignag [31]
3 years ago
15

what is the difference in gravitation potential energy when a 3.5 kg box is raised from a height of 1.2m to height of 4.0m?​

Physics
1 answer:
Whitepunk [10]3 years ago
6 0

Answer: 96.138 J

Explanation:

Ep= mGh (on 1.2 m high)

Ep= 3.5kg*9.81*1.2

Ep1=41.202 J

Ep=mGh (on 4m high)

Ep=3.5*9.81*4

Ep2=137.34J

Ep= Ep2-Ep1

Ep=96.138J

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Answer with Explanation:

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The 5-lb collar is released from rest at A and travels along the frictionless guide. Determine the speed of the collar when it s
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Answer:

Explanation:

Stiffness of spring k equal to 4 lb/ft.

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Weight of the collar W is 15lb

Radius of curvature of curve guide is 1 feet

length of vertical rod is 1.5 feet

Initial speed of collar when released from rest at A is 0 feet per seconds  

use the energy conservation equation

 P_A +K_A=P_B+K_B

Estimate the potential energy , component as position B as below

P_A=Wh_1+\frac{1}{2} ks^2_1\\\\=5\times(1.5+1)+\frac{1}{2} \times 4 \times(1.5+1-0.5)^2\\\\=20.5lb.ft

Estimate the kinetic energy , component as position A as below

K_A=\frac{1}{2} \frac{W}{g} V^2_1\\\\=\frac{1}{2} \frac{5}{32.2} \times0^2\\\\=0lb.ft

Estimate the kinetic energy , component as position B as below

K_A=\frac{1}{2} \frac{W}{g} V^2_2\\\\=\frac{1}{2} \frac{5}{32.2} \times V^2_2\\\\=00777V^2_2

Substitute 20.5lb- ft for P_A

0.5lb-ft for P_B

0lb -ft for K_A

0.0777V_2^2 for K_B

20.5+0=0.5+0.777V^2_2\\\\V^2_2=257.6\\\\V_2=16.05

= 16.05ft/sec

7 0
3 years ago
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