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Physics 2210
Fall 2015
Review for Midterm Exam 2
10/07/2015
Problem 1 (1/3)
A spring of force constant k = 800 N/m and a relaxed
length 𝐿0 = 1.10 m has its upper end fixed/attached to a
pivot in the ceiling. Its free end is attached to a block of
mass m=45.0 kg that sits on a frictionless, horizontal
surface. When the block is directly below the pivot, the
spring is stretched to a length h = 1.20 m. The block is
then pulled left a horizontal distance b = 0.90 m and
released from rest. Calculate
(a) the magnitude of the force exerted by the spring on the block,
(b) the acceleration of block right after being released, and
(c) the speed of block as it passes directly under the pivot again. You can assume
that the block remains on the surface.
Solution: (a)
𝐹𝑠 = 𝑘 ∆𝐿𝑖 ,
∆𝐿𝑖 =𝐿𝑖 -𝐿0 ,
𝐿0 =1.10m, 𝑘=800N/m
𝐿𝑖 = ℎ2 + 𝑏 2 = 1.20m 2 + 0.90m 2 = 1.50m
𝐹𝑠 = 800N/m 1.50m − 1.10m = 320N
Problem 1 (2/3)
Spring: k = 800 N/m, 𝐿0 = 1.10 m attached to a block of mass
m=45.0 kg on a frictionless, horizontal surface. Directly below
pivot, the spring is stretched to h = 1.20 m. The block is then
pulled left a horizontal distance b = 0.90 m and released from
rest. Calculate
(a) the magnitude of the force exerted by the spring on the block,
(b) the acceleration of block right after being released, and
(c) the speed of block as it passes directly under the pivot again
Solution: (b), we had 𝐹𝑠 =320N
If the block stays on the surface then its acceleration is
purely to the right, due to the x-component of the net
force on it. 𝑎𝑦 = 0, and
𝐹𝑥 = 𝐹𝑠 cos 𝜙 = 𝑚𝑎𝑥 ,
𝜙 = tan−1 ℎ⁄𝑏 = tan−1 1.20m⁄0.90m
= tan−1 1.333 = 53.13°
320N cos 53.13°
𝐹𝑠 sin 𝜙
𝑎𝑥 =
=
= 4.27 m⁄s 2
𝑚
45kg
𝑎⃗ = 4.27 m⁄s 2 𝐢̂ or 4.27 m⁄s 2 to the right
Problem 1 (3/3)
Spring: k = 800 N/m, 𝐿0 = 1.10 m attached to a block of mass
m=45.0 kg on a frictionless, horizontal surface. Directly below
pivot, the spring is stretched to h = 1.20 m. The block is then
pulled left a horizontal distance b = 0.90 m and released from
rest. Calculate
(c) the speed of block as it passes directly under the pivot again
Solution (c) Normal force does no work (perpendicular to the path), and there is no
friction. There is no change in height  gravity does no work. Only the spring
force is involved and it is a conservative force.  Total energy is conserved:
𝐸𝑖 = 𝐾𝑖 + 𝑈𝑖 = 12𝑚𝑣𝑖 2 + 12𝑘 Δ𝐿𝑖 2 = 0 + 12𝑘 Δ𝐿𝑖 2 (initially at rest:𝑣𝑖 =0)
𝐸𝑓 = 𝐾𝑓 + 𝑈𝑓 = 12𝑚𝑣𝑓 2 + 12𝑘 Δ𝐿𝑓
𝑚𝑣𝑓 2 + 12𝑘 Δ𝐿𝑓
1
2
2
= 12𝑘 Δ𝐿𝑖
2
2
, and 𝐸𝑓 = 𝐸𝑖
 𝑚𝑣𝑓 2 = 𝑘 Δ𝐿𝑖
2
− Δ𝐿𝑓
2
From part (a) Δ𝐿𝑖 =0.40m and Δ𝐿𝑓 = ℎ − 𝐿0 = 1.20m − 1.10m = 0.10m
2 − 0.10m 2
⁄
800
N
m
0.40m
= 2.667 m2 ⁄s 2
𝑣𝑓 2 =
45.0 kg
𝑣𝑓 = 1.63 m⁄s
Problem 2 (1/3)
A block in the figure has mass 𝑚1 = 6.10 kg.
The coefficient of static friction between the
block and the horizontal desk top is 𝜇𝑠 = 0.40.
The block is attached by a cord to a cowbell of
mass 𝑚2 = 1.25 kg. The cowbell is attached
by a second cord to a hook on the wall. The
first cord is horizontal, and the second cord at
an angle of 𝜃 = 40.0° from the horizontal.
(a) The system is in equilibrium as shown. Calculate the tension 𝑻𝟐 on the block.
(b) Calculate the magnitude of the friction force exerted by the desk on the block.
(c) How much additional mass can be added to the cowbell (i.e. to 𝑚2 ) and still
keep the system in equilibrium?
Solution: (a) consider the freebody diagram of the cowbell.
The system is at equilibrium acceleration of the cowbell is zero.
We start with the y-component of the net force on the cowbell:
𝐹2𝑦 = 𝑇2 sin 𝜃 − 𝑚2 𝑔 = 𝑚2 𝑎2𝑦 = 0 → 𝑇2 sin 𝜃 = 𝑚2 𝑔
𝑚2 𝑔
1.25 kg 9.8 m⁄s 2
→ 𝑇2 =
=
= 19.1 N
sin 𝜃
sin 40.0°
Problem 2 (1/3)
𝑚1 = 6.10 kg. 𝜇𝑠 = 0.40. 𝑚2 = 1.25 kg.
𝜃 = 40.0° from the horizontal.
(a) The system is in equilibrium as shown. Calculate the tension 𝑇2 on the block.
(b) Calculate the magnitude of the friction force exerted by the desk on the block.
(c) How much additional mass can be added to the cowbell (i.e. to 𝑚2 ) and still
keep the system in equilibrium?
Solution: (b) Looking again at the free-body diagram:
The x-component of the net force on the cowbell:
𝐹2𝑥 = 𝑇2 cos 𝜃 − 𝑇1 = 𝑚2 𝑎2𝑥 = 0
𝑚2 𝑔
→ 𝑇1 = 𝑇2 cos 𝜃 → 𝑇1 =
cos 𝜃 = 𝑚2 𝑔 co𝑡 𝜃
sin 𝜃
Now on to the free-body diagram on the block: x-component of net
force: 𝐹1𝑥 = 𝑇1 − 𝑓𝑆 = 𝑚1 𝑎1𝑥 = 0 → 𝑓𝑆 = 𝑇1 = 𝑚2 𝑔 co𝑡 𝜃
𝑓𝑆 = 1.25 kg 9.8 m⁄s 2 cot 40.0° = 14.6N
Compare this to maximum static friction force given by 𝑓𝑆𝑀𝑀𝑀 = 𝜇𝑠 𝑁.
Now, the y-component of the net force on the block is:
𝐹1𝑦 = 𝑁 − 𝑚1 𝑔 = 𝑚1 𝑎1𝑦 = 0 → 𝑁 = 𝑚1 𝑔
𝑓𝑆𝑆𝑆𝑆 = 𝜇𝑠 𝑚1 𝑔 = 0.40 6.10 kg 9.8 m⁄s 2 = 23.9N > 14.6N
Problem 2 (1/3)
𝑚1 = 6.10 kg. 𝜇𝑠 = 0.40. 𝑚2 = 1.25 kg.
𝜃 = 40.0° from the horizontal.
(a) The system is in equilibrium as shown. Calculate the tension 𝑇2 on the block.
(b) Calculate the magnitude of the friction force exerted by the desk on the block.
(c) How much additional mass can be added to the cowbell (i.e. to 𝒎𝟐 ) and still
keep the system in equilibrium?
Solution: (c) From part (b) we had 𝑓𝑆 = 𝑇1 = 𝑚2 𝑔 co𝑡 𝜃,
and 𝑓𝑆𝑆𝑆𝑆 = 𝜇𝑠 𝑚1 𝑔 = 23.9N.
So if we were to increase 𝑚2 , then 𝑓𝑆 increases proportionally
So the maximum total 𝑚2 corresponds to the case 𝑓𝑆 = 𝑓𝑆𝑆𝑆𝑆 :
𝑓𝑆𝑆𝑆𝑆 = 𝑚2𝑀𝑀𝑀 𝑔 co𝑡 𝜃
𝑓𝑆𝑆𝑆𝑆
23.9N
tan 𝜃 =
tan 40.0° = 2.05kg
2
⁄
𝑔
9.8 m s
And so the maximum additional mass on the cowbell that still maintains
equilibrium is
∆𝑚2𝑀𝑀𝑀 = 𝑚2𝑀𝑀𝑀 − 𝑚2 = 2.05kg − 1.25kg = 0.80 kg
→ 𝑚2𝑀𝑀𝑀 =
Problem 3 (1/3)
In the figure, the pulley has negligible mass and is
frictionless. Block A has a mass of 2.5 kg, and block B
4.0 kg. The angle of the incline is 𝜃 = 35°. The
coefficient of kinetic friction between the incline and
block A is 𝜇𝑘 = 0.30. The blocks are set in motion
from rest with the cord taut.
(a) Find the magnitude of the friction force exerted by the incline on block A.
After block B falls a distance of 𝐷=45 cm, calculate
𝑇
𝑁
(b) the work done by gravity on block A, and
(c) the total kinetic energy of the two blocks.
𝑓𝑘
You can assume that block A remains on the incline.
𝑚𝐴 𝑔
Solution: (a) Since 𝑚𝐵 > 𝑚𝐴 block B will clearly drop and A will move up the
incline. So the kinetic friction force acts down the incline. For analyzing block A,
we choose +x to be up the incline, and +y perpendicularly out of the incline.
Y-component of net force on A:
→ 𝑁 = 𝑚𝐴 𝑔 cos 𝜃
𝐹𝐴𝐴 = 𝑁 − 𝑚𝐴 𝑔 cos 𝜃 = 𝑚𝐴 𝑎𝐴𝐴 = 0
𝑓𝑘 = 𝜇𝑘 𝑁 = 𝜇𝑘 𝑚𝐴 𝑔 cos 𝜃 = 0.30 2.5kg 9.8 m⁄s 2 cos 35° = 6.02N
Problem 3 (2/3)
Pulley is massless and frictionless. 𝑚𝐴 =2.5 kg, 𝑚𝐵 =4.0 kg.
𝜃 = 35°. 𝜇𝑘 = 0.30 between the incline and block A.
Blocks are set in motion from rest
(a) Find the magnitude of the friction force exerted by the incline on block A.
After block B falls a distance of 𝑫=45 cm, calculate
𝐷 ∆ℎ𝐴
(b) the work done by gravity on block A, and
(c) the total kinetic energy of the two blocks.
𝜽 + 𝟗𝟗𝟗
𝑚 𝑔
Solution: (b)
𝑊𝑔𝑔 = 𝐹⃗𝑔 ∙ ∆𝑟⃗𝐴 = 𝑚𝐴 𝑔𝐷 cos 𝜃 + 90° = 2.5kg 9.8 m⁄s 2
Note:
𝐴
0.45m cos 125°
cos 𝑎 + 𝑏 = cos 𝑎 cos 𝑏 − sin 𝑎 sin 𝑏
→ cos 𝜃 + 90° = cos 𝜃 cos 90° − sin 𝜃 sin 90° = − sin 𝜃
And we actually have cos 125° = −sin35° = −0.574
𝑊𝑔𝑔 = 2.5kg 9.8 m⁄s 2 0.45m −0.574 = −6.32J
Alternately: By the definition of potential energy:
𝑊𝑔𝑔 = −∆𝑈𝑔𝑔 , and ∆𝑈𝑔𝑔 = 𝑚𝐴 𝑔∆ℎ𝐴 . Here block A moved a distance 𝐷 up an
incline and so ∆ℎ𝐴 = 𝐷 sin 𝜃 → 𝑊𝑔𝑔 = −𝑚𝐴 𝑔∆ℎ𝐴 = −𝑚𝐴 𝑔𝐷 sin 𝜃 which is the
same answer we had before.
Problem 3 (3/3)
Pulley is massless and frictionless. 𝑚𝐴 =2.5 kg, 𝑚𝐵 =4.0 kg.
𝜃 = 35°. 𝜇𝑘 = 0.30 between the incline and block A.
Blocks are set in motion from rest
(a) Find the magnitude of the friction force exerted by the incline on block A.
After block B falls a distance of 𝑫=45 cm, calculate
(b) the work done by gravity on block A, and
(c) the total kinetic energy of the two blocks.
Solution: (c) We will consider block A + block B as a single object. Since we have
friction acting on A (the only non-conservative EXTERNAL force), then we have
(unit 09): ∆𝐸 = 𝑊𝑘 = 𝑓⃗𝑘 ∙ ∆𝑟⃗𝐴 = −𝑓𝑘 𝐷 = − 6.02N 0.45m = −2.71J
(by the way, the force of the cord is internal to the system)
Initially, the blocks are at rest: 𝐾𝑖 = 0, we are asked to find 𝐾𝑓 = 𝐾𝑖 + ∆𝐾 = ∆𝐾
But by the definition of total energy: ∆𝐸 = +∆𝑈, and ∆𝑈 = ∆𝑈𝑔𝑔 + ∆𝑈𝑔𝐵
∆𝑈𝑔𝑔 = 𝑚𝐵 𝑔∆ℎ𝐵 = −𝑚𝐵 𝑔𝐷 = − 4.0kg 9.8 m⁄s 2 0.45m = 17.64J
And so 𝐾𝑓 + ∆𝑈𝑔𝑔 + ∆𝑈𝑔𝑔 = 𝑊𝑘
→ 𝐾𝑓 = −∆𝑈𝑔𝑔 − ∆𝑈𝑔𝑔 + 𝑊𝑘 = −6.32J + 17.64J − 2.71J = 8.61J