68
2 Particle Dynamics
Fig. 2.21
Work required to lift the entire hanging part is
W =
dW =
L/3
o
Mg
L
x dx =
MgL
18
2.8 (a) The equations of motion are
Ma = mg − T
(1)
ma = T − Mgμ
(2)
Solving (1) and (2)
a =
(m − μM)g
M + m
=
(0.45 − 0.2 × 2)9.8
2 + 0.45
= 0.2 m/s
2
(b) T = m(g − a) = 0.45(9.8 − 0.2) = 4.32 N
(c) After 2 s, the velocity will be
υ 1 = 0 + at = 0.2 × 2 = 0.4 m/s
1
When the string breaks, the acceleration will be a 1 = −μg = − 0.2×9.8 =
−1.96 m/s 2 and final velocity v 2 = 0:
S =
v 2
2 − v 2
1
2a 1
=
0 − (0.4) 2
(2)(−1.96)
= 0.0408 m = 4.1 cm
2.3.2 Motion on Incline
2.9 (a) Gravitational force down the incline is Mg sin θ . Frictional force up the
incline is μmg cos θ . Net force
2 Particle Dynamics
Fig. 2.21
Work required to lift the entire hanging part is
W =
dW =
L/3
o
Mg
L
x dx =
MgL
18
2.8 (a) The equations of motion are
Ma = mg − T
(1)
ma = T − Mgμ
(2)
Solving (1) and (2)
a =
(m − μM)g
M + m
=
(0.45 − 0.2 × 2)9.8
2 + 0.45
= 0.2 m/s
2
(b) T = m(g − a) = 0.45(9.8 − 0.2) = 4.32 N
(c) After 2 s, the velocity will be
υ 1 = 0 + at = 0.2 × 2 = 0.4 m/s
1
When the string breaks, the acceleration will be a 1 = −μg = − 0.2×9.8 =
−1.96 m/s 2 and final velocity v 2 = 0:
S =
v 2
2 − v 2
1
2a 1
=
0 − (0.4) 2
(2)(−1.96)
= 0.0408 m = 4.1 cm
2.3.2 Motion on Incline
2.9 (a) Gravitational force down the incline is Mg sin θ . Frictional force up the
incline is μmg cos θ . Net force