Phys101
Term: 122
Online HW-Ch06-Lec02
Q1:
Uniform circular motion is the direct consequence of:
A. Newton's third law
B. a force that is always tangent to the path
C. an acceleration tangent to the path
D. a force of constant magnitude that is always directed away from the same
fixed point
E. a force of constant magnitude that is always directed toward the same
fixed point
Ans:
E, the force is centripetal and is always directed towards the fixed center.
Q2:
Block A, with a mass of 10 kg, rests on a 30o incline. The coefficient of kinetic
friction is 0.20. The attached string is parallel to the incline and passes over a
massless, frictionless pulley at the top. Block B, with a mass of 8.0 kg, is attached to
the dangling end of the string. Find the magnitude of the acceleration (in m/s2) of
block B. (Give your answer in three significant figures form)
Ans:
𝐁𝐥𝐨𝐜𝐤 𝐀:
−Mg sin30 − fk + T = Ma
and fk = µk N = µk Mg cos 30
∴ −Mg sin30 − µk Mg cos 30 + T = Ma ⟶ (1)
𝐁𝐥𝐨𝐜𝐤 𝐁:
−T + mg = ma ⟶ (2)
Adding (1)and (2)and solve for a:
−Mg sin30 − µk Mg cos 30 + mg = (m + M)a
⇒a=
√3
�−10 × 9.8 × 0.5 − 0.2 × 10 × 9.8 × 2 + 8 × 9.8�
KFUPM-Physics Department
(10 + 8)
= 0.690 m/s2
1
Phys101
Term: 122
Online HW-Ch06-Lec02
Q3:
One end of a 1.0 m long string is fixed while the other end is attached to a 3.0 kg
stone. The stone swings in a vertical circle. The stone has a speed of 2.0 m/s when it
passes the bottom point. Find the magnitude of the tension force (in N) of the string at
this bottom point. (Give your answer in three significant figures form)
Ans:
At the bottom point; Newton’s 2nd law gives:
T – mg = ma where a =
⇒ T = (g + a)m
v2
22
=
⇒ a = 4 m/s 2
R
1
= (9.8 + 4) × 3 = 41.4 N
KFUPM-Physics Department
2