=========================preview======================
(PHYS011)mt06solu.pdf
Back to PHYS011 Login to download
======================================================
1. (a) Free-body diagrams:
Newtons second law for the horizontal and vertical motion of
m1: T C f1 = m1a
n1 = m1g
f1 = (0.100) n1 = (0.100) m1g
thus T C (0.100) m1g = m1a (1)

m2: F C T C f2 = m2a
n2 = m2g
f2 = (0.100) n2 = (0.100) m2g
thus F C T C (0.100) m2g = m2a (2)

Adding equations (1) and (2), we have
F C (0.100) m1g C (.0100) m2g =(m1 + m2)a
a = F C (0.100)g(m1 + m2) / (m1 + m2)
= 68.0 C (0.100)(9.80)(12.0 + 18.0) / (12.0 + 18.0) = 1.29 m/s2
From (1),
T = m1 [(0.100)g + a] = 12.0 (0.98 + 1.29) = 27.2 N

(b) Free-body diagrams:



Newtons second law for centripetal motion of in a horizontal plane
ball A: TA = mvA2/rA

ball B: TB C TA = mvB2/rB

The angular speed of the rod at A and B are the same, = vA / rA = vB / rB.
vB = (vA) (rB / rA) = 5.00 (0.400/0.800) = 2.50 m/s.

Thus TA = 0.50(5.00)2/ 0.80 = 15.6 N
TB = 0.50 (2.50)2/ 0.40 + 15.6 = 23.4 N




2. (a) The block starts with vi = 0 and xi = C2.0 x10-2 m. Let the speed of the block be vf at the final position xf = 0. Using conservation of energy, the initial energy is just elastic energy Ui = . kxi2, the final energy is kinetic energy Kf = . mvf2. . mvf = . kxi2 =
vf = [(k/m)xi2 ]1/2 = (1.0x103/1.6)1/2 (2.0x10-2) = 0.50 m/s

(b) Now the block has to overcome friction fk and does extra work
(fk .x) = 4.0(0 C 2.0x10-2) , thus the final kinetic energy Kf is
Kf = Ui C fk .x
. mvf 2 = . kxi2 C fk .x
vf = [(k/m)xi2 C 2 fk.x /m ]1/2
= [(1.0x103/1.6) (2.0x10-2)2 C 2(4.0)(2.0x10-2)/1.6]1/2
= (0.25 C 0.10)1/2 = 0.39 m/s

(c) The forces acting on the block are the restoring force kx and the friction fk. Newtons second law for the block is
kx C fk = ma
Note kx decrease as x decreases. The acceleration a will be positive so the speed will increase as long as kx > fk. The acceleration will be negative when kx < fk , and the speed will decrease. The speed of the block will increase to a maximum value when kx = fk before it decrease. The condition for maximum speed is therefore x = fk / k. This can also be obtained by energy consideration by maximizing the kinetic energy.
x = fk / k = 4.0 / 1.0x103 = 4.0x10-3 m

(d) Same as in (c) except fk = 10.0 N, x = 10.0/ 1.0x103 = 1.0x10-2 m




3. We have the total mechanical energy E is the sum of its potential U and kinetic energy K. U(x) + K(x) = E .


(a) E = 12 J, from the graph, U(2.0) = 8.0 J,
K(2.0) = 4.0 J
= . (0.50)v(2.0)2
v(2.0) = 4.0 m/s



(b) K(x) will be maximum when U(x) is a minimum. From the graph, U(1.0) = 0, U(4.0) = 0.
Thus the maximum kinetic energy, K(1.0) = K(4.0) = 12 J.. The maximum speed vmax is given by
. (0.50) vmax2 = 12
vmax = 6.9 m/s



(c) At the a turning point