The potential energy of a 0.20-kg particle moving along the X axis is given by U

Question

The potential energy of a 0.20-kg particle moving along the X axis is given by U(x) = (8.0 j/m^2) x^2 - (2.0 J/m^4) x^4. When the particle is at x = 1.0 m, find the magnitude of its acceleration. Blocks A and B are moving toward each other. A has a mass of 2.0 kg and a velocity of 50 m s. while B has a mass of 4.0 kg and a velocity of 25 m/s. They suffer a completely inelastic collision. Find the kinetic energy lost during the collision. A wheel starts from rest and has an angular acceleration that is given by a(t) = (6.0 rad/s^4) t^2. Find the time it takes to make 10 rev. A wheel initially has an angular velocity of 18 rad s but it is slowing at a rate of 2.0 rad/s^1. By the time it stops find the angle it will have turned through. A force F = 4.2 N i + 3.7 N J + 1.2 N k acts on a particle located at x = 3.3 m. What is the torque on the particle around the origin?

Explanation / Answer

3) U(x) = 8 x^2 + 2x^4
at x = 1
U(1) = 10 J, KE (1) = 0.5 m v^2 = 0.5*0.20*25 = 2.5 J
Total mech energy = T = 10 + 2.5 = 12.5 J = remains conserved

x = 0, P(0) = 0, KE(0) = 0.5*0.2 *v^2
energy conservation gives
0 + 0.5*0.20* v^2 = 12.5
v^2 = 125
v = 11.48 m/s

b) Force F = - dU/dx = - 16 x - 8 x^3
at x = 1, force acting on particle
F(1) = - 24 N
m a = - 24
a = acceleration = - 24/0.20 = - 120 m/s^2

4) As the collision is completely inelastic, the two blocks stick together after impact. Therefore we can treat them as a single block of mass 6.0 kg after impact. We first need to find the velocity of the 6.0 kg block after impact, and we need to do this using conservation of momentum: Momentum is always conserved in a collision. And momentum is mass times velocity.

Momentum of Block A before
= (2.0 kg)(50 m/s) = 100 kg m/s

Momentum of block B before
= (4.0 kg)(-25 m/s) = -100 kg m/s

Total momentum before collision
= 100 kg m/s + -100 kg m/s
= 0 kg m/s.

Therefore the total momentum after the collision must also be 0 kg m/s, which means the velocity of the 6.0 kg block after impact must be 0 m/s.

As kinetic energy is given by K = 1/2 mv², this means the kinetic energy after the collision is 0 J, and therefore all kinetic energy is lost. So the total kinetic energy lost is the sum of the kinetic energies of blocks A and B before the collision.

Kinetic energy of block A
= 1/2 (2.0 kg)(50 m/s)²
= 2500 J

Kinetic energy of block B
= 1/2 (4.0 kg)(25 m/s)²
= 1250 J.

Kinetic energy lost
= 2500 J + 1250 J
= 3750 J.

5)think this problem can easily be done through integration since you know that the derivative of angular velocity is angular acceleration and the derivative of angular position is angular velocity. So therefore:

omega(t) = 2t^3 rad/s + C

Since wheel starts from rest, C = 0

theta(t) = (t^4)/2 + D

D = 0 since the wheel has not started rotating at time t = 0.

Therefore time taken = 62.83 rad = t^ 4/2

Timetaken = 3.3 sec

6 ) just as linear motion can be described by x(t) = x0 + v0*t + 1/2*a*t^2, we can use the equation

phi(t) = phi0 + omega0 * t + 1/2 * alpha * t^2

for the angular position (phi) as a function of time (t), where phi0 is the initial angular position, omega0 is the initial angular velocity, and alpha is the angular acceleration.

We have: phi0 = 0, omega0 = 18rad/s, and alpha = -2 rad/s/s.

Note that because the angular velocity is zero after 9 seconds ( 18 rad/s / (2 rad/s/s) ), we have

phi( 9 s) = 0 + 18 rad/s * 9 s - 1rad/s^2 * (9s)^2 = 81 rad.

In revolutions, this is 81 rad / (2 pi rad/revolution) ~ 12.9 revolution

7) F = 4.2i + 3.7j + 1.2k N
r = 3.3i + 0j + 0k m

Torque = F cross r = 12k N*m

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