Generally, I Work Hard by Principle - Energy Problem Solutions

Some solutions to tide you over until class:

From Thursday's handout:

4. Your conservation of energy equation should be:
0 + m₂g*h₂ = .5*(m₁ + m₂ + m₃)*v² + m₂g*h₂ + m₁g*h₂ - m₃g*h₂

This is with h = 4 meters, and v = final speed of all three blocks.

Solving, you should obtain that v = 5.112 m/s.

5. You can set your initial position to be when the block is compressed against the spring, and your final at the max height on the incline. You don't have to find the speed in the middle, though it is ok to do so.

.5*k*x² = m*g*h

h = 16.3 cm

6. You can solve this using a free body diagram, obtaining that T = mg sin(30) = 24.5 N

The other way to do this is using the generalized work-energy principle. The work done by tension here counts as W_ncf since tension is NOT a conservative force.

W_ncf = K_f - K₀ + U_f - U₀
T * x * cos(1) = (0 since constant speed) + mg*x*sin(30)
T = mg sin(30) = 24.5 N

Either method works!

7. It is reasonable to assume (since otherwise there would not be enough information) that the force given is the net force acting on the object.
a) At x = 8 meters, F_x,net = 3 N. This means the acceleration a = Fnet/m = 1 m/s².

b) The work is the area under the graph = .5*(5 m + 15 m)* 3 N = 30 J

c) W_net = Kf - Ko = .5*m*v^2 - 0 so v = (2*Wnet/m)^(1/2) = 4.47 m/s

d) Wncf = Kf - Ko with Kf = 0, and Ko = the answer from b).

Using a FBD, it can be shown that Wncf = mu*mg*delta_x*(-1) = -.5*m*v_f^2.

Thus mu = mv_f^2/(2*g*delta_x) = .204

Problems from Friday:
Worksheet:
Problem 4.
a) v0 = (gR)^(1/2)
b) v = (5gR)^(1/2)
c) Using Wnet = delta_K:
mu*mg*D*(-1) = - .5*m*v^2

mu = 5R/2D

Problem 5.
Wncf = delta_K + delta_U
delta_K = 0 since it is at rest at the start and finish.
delta_U = mg(H2 - H1)

Substituting:
F_fric*delta_x*(-1) = mg(H2 - H1) with H2 = 34 m, and H1 = 38 m.
F_fric = 588 N

Book HW:
Questions:
12. The initial speed of all three balloons is the same. The initial height of all three balloons is the same. The final height of all three balloons is the same.

As a result, since only gravity does work, energy is conserved, and the final speed will be the same in all three situations.

17. At the maximum height, all of the energy is gravitational potential energy. At the moment just before the spring in the pogo stick is compressed, all of the energy is kinetic. At the bottom of the jump, all of the energy is stored in the spring as elastic potential energy. Since gravity and the elastic force are conservative forces, the total energy remains constant.

Problems:
39.
a) v = 9.154 m/s (don't forget to use the initial speed of 5 m/s!)
b) .5*m*v^2 + 0 = .5*k*x^2 + m*g*(-x)
x = 0.362 m

43.
a)
.5*m*v^2 + 0 = .5*k*h^2 + m*g*(-h)
solving, v = 8.03 m/s
b) 3.290 meters

44.
This is the same as the problem we did in class: h = 5/2 R

45.
a) E = K + U = (1/2)mv^2 + 1/2*k*x^2

76.
a) v = (2gL)^2
b) Setting U_g = 0 at the bottom of the swing:
mgL + 0 = .5*m*v_f^2 + mg(2*.2L)
v_f = (1.2*gL)^(1/2)

85.
a) We can use conservation of energy since only gravity is doing work.
.5*mv_o^2 + 0 = 0 + mgL(1-cos(theta))
cos(theta) = 1 - v_o^2/(2*g*L)
theta = 29.3 degrees

b) To solve this, we need to make the observation that Tarzan is moving in circular motion. At the top of the swing, the y-axis must be pointed towards the center of the circle.
T - mg*cos(theta) = 0 (since v = 0, the centripetal acceleration is zero)
T = 640 N

c) Tension will be the greatest at the bottom when Tarzan first grabs the rope. (This is the point when v is greatest, and therefore the centripetal acceleration as well.)

T = mg + mv^2/L = 922.5 N