1. use x-x0 = v0t + 1/2 a t².

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_i².
E_f = m g 100 + 1/2 m v_f² + 1000 J. Solve for v_f.

8. Use K_tot = 1/2 M V² + 1/2 I omega² 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 v² + 1/2 I_sphere omega². Use omega = V/R and solve for V.

1. Again, conserve energy. E_i = 1/2 M v² + 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 u² + 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/r² 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:
N₁ = 3M g cos(phi)
3M g sin(phi) - T₁ = 3M a
M block:
N₂ = M g cos(phi)
T₂ - M g sin(phi) = M a
pulley:
(T₁-T₂)*r = I alpha
alpha = a/r.
Solve for a:
a = 2 M g sin(phi)/ [4 M + I/r²].