Question

A 3.0 kg object is moving along the x-axis in a region where its potential energy as a function of x is given as U(x) = 4.0x2 , where U is in joules and x is in meters. When the object passes the point x = -0.50 m, its velocity is +2.0 m/s. All forces acting on the object are conservative. Calculate the total mechanical energy of the object Calculate the x-coordinate of any points at which the object has zero kinetic energy. Calculate the magnitude of the momentum of the object at x = 0.60 m. Calculate the magnitude of the acceleration of the object as it passes x = 0.60 m. On the axes below, sketch graphs of the object’s position x versus time t and kinetic energy K versus time t. Assume that x = 0 at time t = 0 . The two graphs should cover the same time interval and use the same scale on the horizontal axes.

Answer 1

Answer:

a) $E_{tot} = 7\,J$, b) $x = \pm\sqrt{\frac{7}{4} }$c) $p = 5.775\,\frac{kg\cdot m}{s}$

Explanation:

a) The total energy of the object is:

$E_{tot} = U + K$

$E_{tot} = 4\cdot x^{2} + \frac{1}{2}\cdot m \cdot v^{2}$

$E_{tot} = 4\cdot (-0.50\,m)^{2} + \frac{1}{2}\cdot (3\,kg)\cdot (2\,\frac{m}{s} )^{2}$

$E_{tot} = 7\,J$

b) The total energy of the object is:

$E_{tot} = U$

$7\,J = 4\cdot x^{2}$

$x = \pm\sqrt{\frac{7}{4} }$

c) The speed of the object is clear in the total energy expression:

$E_{total} = U + K$

$K = E_{total}-U$

$\frac{1}{2}\cdot m \cdot v^{2} = E_{total} - 4\cdot x^{2}$

$v^{2} = \frac{2\cdot (E_{total}-4\cdot x^{2})}{m}$

$v = \sqrt{\frac{2\cdot (E_{total}-4\cdot x^{2})}{m} }$

$v = \sqrt{\frac{2\cdot [7\,J- 4\cdot (0.6\,m)^{2}]}{3\,kg} }$

$v \approx 1.925\,\frac{m}{s}$

The magnitude of the momentum is:

$p = (3\,kg)\cdot (1.925\,\frac{m}{s} )$

$p = 5.775\,\frac{kg\cdot m}{s}$

d) Before calculating the acceleration experimented by the object, it is required to determine the net force exerted on it. There is a relationship between potential energy and net force:

$F = -\frac{dU}{dx}$

$F = -8\cdot x$

Acceleration experimented by the object is:

$a = -\frac{8\cdot x}{m}$

$a = -\frac{8\cdot (0.6\,m)}{3\,kg}$

$a = -1.6\,\frac{m}{s^{2}}$

e) The position of the object versus time is found by solving the following differential equation:

$\frac{d^{2}x}{dt} +\frac{8\cdot x}{m} = 0$

$s^{2}\cdot X(s)- s\cdot v(0) - x(0) + \frac{8}{m}\cdot X(s) = 0$

$(s^{2} + \frac{8}{m})\cdot X(s) = s\cdot v(0)+x(0)$

$X(s) = \frac{s\cdot v(0)+x(0)}{(s^{2}+\frac{8}{m} )}$

$X(s) = v(0)\cdot \frac{s}{s^{2}+\frac{8}{m} } +\frac{m\cdot x(0)}{8} \cdot \frac{\frac{8}{m}}{s^{2}+\frac{8}{m}}$

$x(t) = v(0) \cdot \cos \left(\frac{8}{m}\cdot t \right)+\frac{m\cdot x(0)}{8}\cdot \sin \left(\frac{8}{m}\cdot t \right)$

The velocity is obtained by deriving the previous expression:

$v(t) = -\frac{8\cdot v(0)}{m}\cdot \sin \left(\frac{8}{m}\cdot t \right)+x(0)\cdot \cos \left(\frac{8}{m}\cdot t \right)$

Speed of the object at $x = 0$ is:

$v = \sqrt{\frac{2\cdot (E_{total}-4\cdot x^{2})}{m} }$

$v = \sqrt{\frac{2\cdot [7\,J- 4\cdot (0\,m)^{2}]}{3\,kg} }$

$v \approx 2.160\,\frac{m}{s}$

The equation of the motion are:

$x(t) = 2.160\cdot \cos \left(2.667\cdot t \right)$

$v(t) = -5.76\cdot \sin (2.667\cdot t)$

The expression for the kinetic energy of the object is:

$K = \frac{1}{2}\cdot m \cdot v^{2}$

$K = \frac{1}{2}\cdot (3\,kg)\cdot [33.178\cdot \sin^{2}(2.667\cdot t)]$

$K = 49.767\cdot \sin^{2}(2.667\cdot t)$

Graphics are included below as attachments. (Position versus time, kinetic energy vs time).

Answer 2

Following are the responses to the given points:

Given:

object mass $(m)= 3.0\ kg$

Potential energy $U(x)=4.0\ x^2$  

$\to x=-0.5 \ m\\\\ \to velocity \ (v)=2.0 \ \frac{m}{s}$

Solution:

For point (a)  

Total Energy $(TE) = PE + KE$

$\to PE = 4.0 (0.5 m)^2 = 1\ J\\\\ \to KE = \frac{1}{2} mv^2 = 0.5 \times 3.0\ kg (2.0 \frac{m}{s})^2 = 6\ J \\\\\to TE = 7\ J$

For point (b)  

If an object's potential energy is 7 J, it has 0 kinetic energy.

$\to U(x) = 4x^2 = 7\ J \\\\\to x=+1.2 \ m, -1.2 \ m$

For point (c)  

$\to x=0.60\ m \\\\ \to TE=4x^2 + \frac{1}{2}mv^2 = 7\ J\\\\ \to 4(0.60)^2 +0.5 \times 3.0 \ kg \times v^2 = 7\ J \\\\ \to v=1.92 \ \frac{m}{s}\\\\ \to Momentum (p) = mv \\\\\to p= 3.0\ kg \times 1.92 \ \frac{m}{s}\\\\ \to p= 5.76 kg \ \frac{m}{s}$

For point (d)

$\to x_1= 0.6 \ m\\\\ \to v_1 = 1.92 \frac{m}{s}\\\\ \to x_2 = 1.2\ m$ the velocity is found by  

$\to 4x^2 + \frac{1}{2} mv^2= 7\ J \\\\ \to 4(1.2)^2 +0.5 \times 3.0\ kg \times v^2 = 7\ J \\\\ \to v_2 = 0.9 \ \frac{m}{s}$

Calculating the acceleration:

$\to V^2_{2}-v^2_{1}= 2a(x_2-x_1) \\\\\to (0.9 \ \frac{m}{s})^2 - (1.92\ \frac{m}{s})^2 = 2 a(1.2\ m - 0.6\ m) \\\\\to a= -2.4 \frac{m}{s^2}$

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