A particle with a mass m = 4.00 kg is moving along the x axis under the influenc

Question

A particle with a mass m = 4.00 kg is moving along the x axis under the influence of the potential energy function U(x) = (2.00 J/m2)x2 32.0 J. If the particle is released from rest at the position x = 6.40 m, determine the following. (The sign is important. Be sure not to round intermediate calculations.)

(a) total mechanical energy of the particle at any position
J

(b) potential energy of the particle at the position

x = 4.60 m

J

(c) kinetic energy of the particle at the position

x = 4.60 m

J

(d) maximum speed of the particle as it travels along the x axis
m/s

(e) acceleration of the particle at the position

x = 4.60 m

(Indicate the direction with the sign of your answer.)
m/s2

(f) maximum position the particle may obtain along the x axis
m

Explanation / Answer

Given that,

U = 2x^2 - 32 J

at x = 6.4 m, particle is at rest.

so, v = 0

KE = 0

PE at x = 6.4 m,

U = 2*6.4^2 - 32 = 49.92 J

Total  mechanical energy of the particle,

TE = PE + KE

TE = 49.92 + 0

TE = 49.92 J

(b)

at the position x = 4.60 m,

U = 2*(4.6)^2 - 32

U = 10.32 J

(c)

At x = 4.6 m,

From law of energy conservation,

TE = KE + PE

KE = 49.92 - 10.32

KE = 39.6 J

(d)

Speed will be maximum when KE is maximum.

for KEmax, PE should be minimum.

PEmin = -32 J

KEmax + PEmin = TE

(1/2)mv^2 -32 = 49.92

(1/2)*4*v^2 = 81.92

vmax = 6.4 m/s

(e)

F = -dU / dx

F = -d(2x^2 - 32) / dx = -4x

at x = 4.5 m,

F = -4*4.6 = -18.4

a = F / m

a = -18.4 / 4

a = -4.6 m/s^2

(f)

At maximum position , PE is maximum.

PEmax = 49.92 J

2*x^2 - 32 = 49.92

x = 6.4 m

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