Problem#1
In the system shown in Figure 1, a horizontal force Fx acts on the 8.00-kg object. The horizontal surface is frictionless. (a) For what values of Fx does the 2.00-kg object accelerate upward? (b) For what values of Fx is the tension in the cord zero? (c) Plot the acceleration of the 8.00-kg object versus Fx. Include values of Fx from –100 N to +100 N.
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| Fig.1 |
Answer:
Forces acting on 2.00 kg block:
T - m1g = m1a
Forces acting on 2.00 kg block:
FX – T = m2a
(a) Forces acting on 2.00 kg block:
a = (Fx – m1g)/(m1 + m2)
then, a > 0 for Fx > m1g = 19.6 N
(b) Forces acting on 2.00 kg block:
T = m1(Fx + m2g)/(m1 + m2)
Then, T = 0 for Fx ≤ -m2g = -78.4 N
(c)
Fx, N –100 –78.4 –50.0 0 50.0 100
ax, m/s2 –12.5 –9.80 –6.96 –1.96 3.04 8.04
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Problem#2
A frictionless plane is 10.0 m long and inclined at 35.0°. A sled starts at the bottom with an initial speed of 5.00 m/s up the incline. When it reaches the point at which it momentarily stops, a second sled is released from the top of this incline with an initial speed vi. Both sleds reach the bottom of the incline at the same moment. (a) Determine the distance that the first sled traveled up the incline. (b) Determine the initial speed of the second sled.
Answer:
(a) For force components along the incline, with the upward direction taken as positive,
∑Fx = max
–mg sin θ = max
ax = –gsin θ
ax = –(9.80 m/s2) sin 35.00 = –5.62 m/s2
For the upward motion,
vxf2 = vxi2 + 2a(xf – xi)
0 = (5 m/s)2 + 2(–5.62 m/s2)(xf – 0)
xf = 2.22 m
(b) The time to slide down is given by
xf = xi + vxit + ½ axt2
0 = 2.22 m + 0 + ½ (–5.62 m/s2)t2
t = 0.890 s
For the second particle,
xf = xi + vxit + ½ axt2
0 = 10 m + vix(0.890 s) + ½ (–5.62 m/s2)(0.890 s)2
vxi = –8.74 m/s
then, speed = 8.74 m/s
Problem#3
A 72.0-kg man stands on a spring scale in an elevator. Starting from rest, the elevator ascends, attaining its maximum speed of 1.20 m/s in 0.800 s. It travels with this constant speed for the next 5.00 s. The elevator then undergoes a uniform acceleration in the negative y direction for 1.50 s and comes to rest. What does the spring scale register (a) before the elevator starts to move? (b) during the first 0.800 s? (c) while the elevator is traveling at constant speed? (d) during the time it is slowing down?
Answer:
First, we will compute the needed accelerations:
(1) Before it starts to move: ay = 0
(2) During the first 0.800 s:
vyf = vyi + ayt
1.2 m/s = 0 + ay(0.800 s)
ay = 1.50 m/s2
(3) While moving at constant velocity: ay = 0
(4) During the last 1.50 s:
vyf = vyi + ayt
0 = 1.2 m/s + ay(1.50 s)
ay = –0.80 m/s2
Newton’s second law is:
∑Fy = ma
+S – (72.0 kg)(9.80 m/s2) = (72.0 kg)ay
S = 706 N + (72.0 kg)ay
(a) When ay = 0 , S = 706 N .
(b) When ay =1.50 m/s2, S = 814 N .
(c) When ay = 0 , S = 706 N .
(d) When ay = –0.800 m/s2, S = 648 N
Problem#4
An object of mass m1 on a frictionless horizontal table is connected to an object of mass m2 through a very light pulley P1 and a light fixed pulley P2 as shown in Figure 2. (a) If a1 and a2 are the accelerations of m1 and m2, respectively, what is the relation between these accelerations? Express
(b) the tensions in the strings and (c) the accelerations a1 and a2 in terms of the masses m1 and m2, and g.
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| Fig.2 |
Answer:
(a) Pulley P1 has acceleration a2 .
Since m1 moves twice the distance P1 moves in the same
time, m1 has twice the acceleration of P1 , i.e., a1 = 2a2
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(b) From the figure, and using
∑F = ma
m2g – T2 = m2a (1)
T1 = m1a1 (2)
T2 - 2T1 = 0 (3)
Equation (1) becomes m2g - 2T1 = m2a2. This equation combined with Equation (2) yields
{2m1 + m2/2}T1/m1 = m2g
Then
T1 = 2m1m2g/(4m1 + m2)
And
T2 = 4m1m2/(4m1 + m2)
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