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• 1. A ball of mass m is suspended from two strings of unequal length
as shown above. The magnitudes of the tensions T1 and T2 in the
strings must satisfy which of the following relations?
• (A) Tl = T2 (B) T1 > T2 (C) T1 < T2 (D) Tl + T2 = mg
Answer
• As T2 is more vertical, it is supporting more of the
weight of the ball. The horizontal components of T1
and T2 are equal.
•C
Questions 4-5: A 2-kilogram block slides down a 30° incline as shown above with an acceleration of 2
meters per second squared.
4) 1984.6
Which of the following diagrams best represents the gravitational force W. the frictional
force f, and the normal force N that act on the block?
Answer
• 2. Normal force is perpendicular to the incline, friction
acts up, parallel to the incline (opposite the motion of
the block), gravity acts straight down.
•E
Questions 4-5: A 2-kilogram block slides down a 30° incline as shown above with an acceleration of 2
meters per second squared.
Which of the following correctly indicates the magnitudes of the forces
acting up and down the incline?
• (A) 20 N down the plane, 16 N up the plane
• (B) 4 N down the plane, 4 N up the plane
• (C) 0 N down the plane, 4 N up the plane
• (D) 10 N down the plane, 6 N up the plane
Answer
• The component of the weight down the plane is 20 N sin 30. The net
force is 4 N, so the friction force up the plane must be 4 N less than
10 N.
•D
A plane 5 meters in length is inclined at an angle of 37°, as shown above. A
block of weight 20 N is placed at the top of the plane and allowed to slide
down.
The magnitude of the normal force exerted on the block by the plane is
• (A) greater than 20 N
• (B) greater than zero but less than 20 N
• (C) equal to 20 N
• (D) zero
Answer
• The weight component perpendicular to the plane is 20 N sin 37o. To
get equilibrium perpendicular to the plane, the normal force must
equal this weight component, which must be less than 20 N.
•B
• a.
• b. f = μN where N = m1g cos θ gives μ = f /(m1g cos θ )
• c. constant velocity means ΣF = 0 where Σ Fexternal = m1g sin θ + m2g
sin θ – f – 2f – Mg = 0
• solving for M gives M = (m1 + m2) sin θ – (3f)/g
• d. Applying Newton’s second law to block 1 gives Σ F = m1g sin θ – f =
m1a which gives a = g sin θ – f/m1