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. Ei = m g 200 + 1/2 m vi2.
Ef = m g 100 + 1/2 m vf2 + 1000 J. Solve for vf.
8. Use Ktot = 1/2 M V2 + 1/2 I omega2 and omega = v/R for rolling to get (d).
10. Conserve energy: Call the radius of the hemisphere R and the radius of the ball r. Then
Ei = M g (R+r). This must equal Ef which is
M g (R+r)(1 - cos(phi)) + 1/2 M v2 + 1/2 Isphere omega2. Use omega = V/R and solve for V.
1. Again, conserve energy. Ei = 1/2 M v2 + 1/2 Isphere (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
Ef = 1/2 M u2 + 1/2 Isphere [u/(R-h)]2 + 0. Solve for u.
2. Solve this with a free body diagram :
linear forces: mg - T = ma
angular forces: Tr = Iyoyo alpha
alpha = a/r
So a = g/(1 + I/r2 m) and finally use I = Isphere(R) + Icylinder(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.
Set up free body diagrams for the two blocks and the pulley. Put the 3M mass on the right side of
the triangle. Then
N1 = 3M g cos(phi)
3M g sin(phi) - T1 = 3M a
N2 = M g cos(phi)
T2 - M g sin(phi) = M a
(T1-T2)*r = I alpha
alpha = a/r.
Solve for a:
a = 2 M g sin(phi)/ [4 M + I/r2].