Block A has a mass of 12 kg. The coefficient of kinetic friction is 0.15. The attached string is parallel to the incline and passes over a massless, frictionless pulley at the top and is attached to block B as shown. The incline makes angle of 30o with the horizontal.
3. If the tension in the string is 40N and block
A is moving down the 30o incline, the acceleration
of block A is closest to:
4. If the tension in the string is 40N and block
A is moving down the 30o incline,
the mass of block B is closest to:
5. If block A is moving up the 30o
incline
at constant velocity, the mass of block B is closest to:
A block of mass 12 kg is pulled in the direction shown in the figure across a rough surface. Force T and the angle q are adjustable and can be changed.
6. If q =
0o and T = 35 N when the block is on the verge of sliding, the
coefficient of static friction is closest to:
7. If we change the angle q
to 30o and set force T = 38 N, the normal force
on the 12-kg block B is closest to:
8. Keep the angle q
= 30o and force T = 38 N. Take the coefficient of
kinetic friction (mk)
between the block and the rough surface to be 0.22, the magnitude of the
frictional force is closest to:
The 5-kg block shown in the figure is attached to a vertical rod by two strings. When the 5-kg mass rotates in a horizontal circle with a speed of 6 m/s, the strings are extended as shown.
9. The time it takes for the 5-kg mass to make one
revolution is closest to:
10. The centripetal acceleration of the 5-kg mass
is closest to (Use the coordinate axis shown to determine the direction
of the centripetal acceleration):
11. The tension in the upper string is closest
to:
A man pushes a 10-kg box with a force of 100 N. For
this series of problems take g = 10 m/s2.
The box moves 20 m across the floor with a constant velocity of 6 m/s.
12. The work done by man on the box is closest to:
13. The work done by friction force on the box is closest
to:
14. The total work done on the box is closest to:
After traveling the 20 m shown, the man increases his force causing the block to speed up to 20 m/s as it moves an additional 10 m. In this second interval:
15. The change in the kinetic energy of the box
is closest to:
16. The total work done on the box is closest to:
A 0.50-kg block is held at rest against the spring by a 36 N external force. The spring constant (k) is 1,800 N/m. The block is released and has a velocity v1 = 1.2 m/s upon separation from the spring. The top surface is frictionless. The block descends a frictionless ramp to where it has a velocity v2 = m1.8 m/s. The block reaches point A and moves onto a rough section whose coefficient of kinetic friction is 0.30. At point B, the velocity of the block is v3 = 1.4 m/s. The block moves on to point C where it stops.
17. The initial compression of the spring, in mm,
is closest to:
18. The height h between the upper and lower levels,
in mm, is closest to:
19. The work done by friction, between point A
and point B, in SI units, is closest to:
20. The distance x that the block travels between
point A and point C, in SI units, is closest to:
Solutions:
1
list of forces on the first FBD (0,N)
(fk,0) (T,0) (-120sin30), -120cos30)
list of forces on the second FBD +T,
-wB
3
4
From second FBD (knowing that the acceleration is pointing up with
value of +4m/s2 if y axis points up)
T-wB = mBa
T=40N, wB = mBg mB
= T/(g+a)g
(-fk+T-120sin30, N-120cos30)=12 (0,0) = (0,0)
and we have to find T
If block A moves with constant velocity (a = 0) so does the block B
and T-wB= T -mBg = 0
8B
9D
10B ar = v2/r
and points to the center
11
12D
13D
14A if constant velocity
then no gain in Kinetic Energy >> total work must be = 0
15A
16B Total work
done is equal to change in Kinetic Energy so from 15 1.82 *103
17C
18C
19A
read eg. KatC = Kinetic Energy at point C: 
20D
