# Q1: Uniform circular motion is the direct consequence of: A ```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
Ans:
𝐁𝐥𝐨𝐜𝐤 𝐀:
−Mg sin30 − fk + T = Ma
and fk = &micro;k N = &micro;k Mg cos 30
∴ −Mg sin30 − &micro;k Mg cos 30 + T = Ma ⟶ (1)
𝐁𝐥𝐨𝐜𝐤 𝐁:
−T + mg = ma ⟶ (2)
Adding (1)and (2)and solve for a:
−Mg sin30 − &micro;k Mg cos 30 + mg = (m + M)a
⇒a=
√3
�−10 &times; 9.8 &times; 0.5 − 0.2 &times; 10 &times; 9.8 &times; 2 + 8 &times; 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
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) &times; 3 = 41.4 N
KFUPM-Physics Department
2
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