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A 3.0-kilogram object is placed on a frictionless track at point A and released from rest.
(Assume the gravitational potential energy of the system to be zero at point C.)
- 4.0 m
m-3.0 kg
3.0 m
2.0 m
1.0 m
0.0 m
b. Calculate the kinetic energy of the object at point B. [Show all work, including the equation
and substitution with units.] [2]

Respuesta :

ajeigbeibraheem

Answer:

Explanation:

From the given information:

The gravitational potential energy at A is:

[tex]PE_A = mgh[/tex]

GIven that:

mass(m) = 3

height (h) = 3

[tex]PE_A =3\times 9.81 \times 3[/tex]

[tex]\mathbf {PE_A =88.29 \ J }[/tex]

b)

Using the conservation of mechanical  energy:

[tex]PE_A+KE_A = PE_B + KE_B[/tex]

[tex]88.29 +0 = 3(9.8)(1) + KE_B[/tex]

[tex]88.29 = 29.4 + KE_B[/tex]

[tex]KE_B = 88.29 -29.4[/tex]

[tex]\mathbf{ KE_B= 58.89 \ J}[/tex]

mickymike92

From the data provided and the calculations done, the potential energy at A is 88.3 J while the kinetic energy at B is 58.9 J.

What is the Potential energy at A?

The potential energy of a body is given as:

  • PE = mgh

where:

m = mass of the object

g = acceleration due to gravity

h = h

At A:

m= 3.0 kg

g = 9.8 m/s^2

h = 3.0 m

PE = 3 × 9.8 × 3

PE = 88.3 J

What is the kinetic energy at B?

Kinetic energy, KE of a body is given as:

  • KE = 1/2 × mv^2

where:

  • m is mass of the body
  • v is velocity of the body

From the law of conservation of energy, the sum of potential energy and Kinetic energy at every point is equal.

PE + KE at A = PE + KE at B

KE at B = PE + KE at A - PE at B

At A, PE = 88.3 J and KE = 0

At B; h = 1.0 m

PE = 3 × 9.8 × 1

PE = 29.4 J

Therefore, KE at B = 88.3 + 0 - 29.4

KE at B = 58.9 J

Therefore, the potential energy at A is 88.3 J while the kinetic energy at B is 58.9 J.

Learn more about potential energy and Kinetic energy at: https://brainly.com/question/14427111

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