1. use x-x0 = v0t + 1/2 a t^{2}.

2. Use |a| = |b X c| = b c sin(angle between them) = 4*6*sin(135)

3.if the block is not to move when the plane accelerates to the right, it must also accelerate at 3 m/s2. Set up the free body diagram and extract the following equations:

y axis: mg = N cos(theta) + f sin(theta)

x axis: ma = N sin(theta) - f cos(theta)

f = mu N

Substitute for f, eliminate N, and obtain mu = [g/a sin(theta)- cos(theta)]/[sin(theta) + g/a cos(theta)].

4. (c) and possibly (d) and (e).

5. this is a conservation of linear momentum or conservation of the center of mass problem.
If the plane moves to the right at 4 m/s, the block must move to the left at 8*4/2.5 m/s.
(The angle of the plane is irrelevant -- except for determining the *vertical* component
of the velocity).

6. Use P = F*v to solve this. We know P = 50*746 and the force the motor exerts (=F) must equal the force of water friction since the boat has constant speed. Lastly we know v, so divide to get |f| = |F| = 50*746/10.

7. Conserve total energy to solve this problem. E_{i} = m g 200 + 1/2 m v_{i}^{2}.

E_{f} = m g 100 + 1/2 m v_{f}^{2} + 1000 J. Solve for v_{f}.

8. Use K_{tot} = 1/2 M V^{2} + 1/2 I omega^{2} and omega = v/R for rolling
to get (d).

9. (d)

10. Conserve energy: Call the radius of the hemisphere R and the radius of the ball r. Then
E_{i} = M g (R+r). This must equal E_{f} which is

M g (R+r)(1 - cos(phi)) + 1/2 M v^{2} + 1/2 I_{sphere} omega^{2}.
Use omega = V/R and solve for V.

1. Again, conserve energy. E_{i} = 1/2 M v^{2} + 1/2 I_{sphere} (v/R)^{2} + m g h. Here h = 10 cm, M is the mass of the ball, and m is the mass of the block. Now lower
the block to get

E_{f} = 1/2 M u^{2} + 1/2 I_{sphere} [u/(R-h)]^{2} + 0.
Solve for u.

2. Solve this with a free body diagram :

linear forces: mg - T = ma

angular forces: Tr = I_{yoyo} alpha

alpha = a/r

So a = g/(1 + I/r^{2} m) and finally use I = I_{sphere}(R) + I_{cylinder}(r).

3. Use W = integral (tau d theta). In this case tau = 0.8 is a constant so W = tau* delta theta= tau* 20 rad/s * 60 sec = 960 J.

4.
Set up free body diagrams for the two blocks and the pulley. Put the 3M mass on the right side of
the triangle. Then

3M block:

N_{1} = 3M g cos(phi)

3M g sin(phi) - T_{1} = 3M a

M block:

N_{2} = M g cos(phi)

T_{2} - M g sin(phi) = M a

pulley:

(T_{1}-T_{2})*r = I alpha

alpha = a/r.

Solve for a:

a = 2 M g sin(phi)/ [4 M + I/r^{2}].