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A 3.0 kg object is moving along the x-axis in a region where its potential energy as function of x is given as U(X) = 4.0x^2, where U is in joules and x is in meters. When the object passes the point x = -0.40m, its velocity is +3.0 m/s. All forces acting on the object are conservative. (No friction).
a. Calculate the total mechanical energy of the object.
b. Calculate the x-coordinate of any points at which the object has zero kinetic energy
c. Calculate the magnitude of the momentum of the object at x = 0.60m
d. Calculate the magnitude of the acceleration of the object as it passes x = 0.40m
e. What direction is the acceleration when x = 0.40m?

Respuesta :

a) The total mechanical energy of the object is 6.04J.

b) The x-coordinate of any points at which the object has zero kinetic energy is 1.2m.

c) The magnitude of the momentum of the object at x = 0.60m is 5.01m/s.

d) The magnitude of the acceleration of the object as it passes x = 0.40m is  2.822 m/s.

e) The direction is the acceleration when x = 0.40m is -6.556m/s².

When a body's velocity is decreasing, it typically experiences negative acceleration. It moves in the opposite direction to the body's motion.

What would you consider potential energy to be?

Potential energy is a form of stored energy that is dependent on the arrangement of a system's components. When stretched or compressed, a spring has more potential energy. When raised above the ground, a steel ball has more potential energy than when it is brought to rest on the ground.

The following are responses to the points made:

Given,

object mass m = 3kg

Potential energy u(x) = 4x²

x = -0.4m

velocity v = 3m/s.

For point (a) Total Energy = PE + KE

PE = 4×(-0.4)² = 0.64J.

KE = 1/2mv²

KE = 1/2(0.4)(3)(3)²

KE = 5.4J

TE = 6.04J.

For point (b)  

If an object's potential energy is 6.04J, it has 0 kinetic energy.

u(x) = 4x² = 6.04J

x = + 1.2m, - 1.2m

For point (c)

x =  0.60m

TE = 4x² + 1/2mv² = 6.04J

4×0.6² + 1/2 × 3 × v² = 6.04

v² = 6.04/4×0.6² + 1/2 × 3

v² = 6.04/2.16

v² = 2.79

v = 1.67m/s

momentum p = mv

p = 3×1.67

p = 5.01m/s.

For point (d)

x₁ = 0.6m

v₁ =  1.67m/s

x₂ = 1.2m the velocity is found by

4x² + 1/2mv² = 6.04J

4(0.4)² + 1/2 × 3 × v² = 6.04J

v₂ = 6.04/2.14

v₂ = 2.822 m/s.

e) Calculating the acceleration:

v₂² - v₁² = 2a(x₂ - x₁)

2.82² - 1.67² = 2a (1.2-0.4)

a = 7.95-2.788/2(0.8)

a = -6.556m/s².

To know more about potential energy visit:

https://brainly.com/question/15521589

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