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The document presents a series of physics problems and solutions related to force and motion, covering topics such as mass calculation, weight at altitude, Hooke's law, and Newton's laws of motion. Each problem includes calculations and formulas used to derive the answers, demonstrating the application of fundamental physics principles. The solutions provide specific numerical results for various scenarios involving forces, weights, and accelerations.

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9 views3 pages

4 Revised

The document presents a series of physics problems and solutions related to force and motion, covering topics such as mass calculation, weight at altitude, Hooke's law, and Newton's laws of motion. Each problem includes calculations and formulas used to derive the answers, demonstrating the application of fundamental physics principles. The solutions provide specific numerical results for various scenarios involving forces, weights, and accelerations.

Uploaded by

민솔
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© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
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<Chap.

4 Force and Motion: Suggested Problems and Solutions>


4.13, 4.24, 4.35, 4.41, 4.51, 4.58

(4.13)

The mass of plane is:

𝐹𝐹 7.0 × 103 N
𝑚𝑚 = = m
𝑎𝑎 7.7 2
s
∴ 𝑚𝑚 = 9.1 × 102 kg

(4.24)

Weight is given by:

𝑤𝑤 = 𝑚𝑚𝑚𝑚
= (68 kg)(9.81m/s 2 )
= 667.08 N

The apparent weight at the altitude will be:

𝑤𝑤𝑎𝑎𝑎𝑎 = (0.89)𝑚𝑚𝑚𝑚
= (0.89)(68 kg)(9.81 m/s 2 )

∴ 𝑤𝑤𝑎𝑎𝑎𝑎 = 593.70 N

(4.35)

Use Hooke’s law:

𝐹𝐹𝑠𝑠 = −𝑘𝑘𝑘𝑘
= −(340 N/m)(𝑥𝑥 m)

𝑤𝑤𝑓𝑓𝑓𝑓𝑓𝑓ℎ = 𝑚𝑚𝑚𝑚
= (6.7 kg)(9.81 m/s 2 )

|𝐹𝐹𝑠𝑠 | = �𝑤𝑤𝑓𝑓𝑓𝑓𝑓𝑓ℎ �

(6.7 kg)(9.81 m/s 2 )


𝑥𝑥 =
340 N/m

∴ 𝑥𝑥 = 0.19 m (or 19 cm)


(4.41)

Use figure 4.11, the acceleration is given by:

𝑇𝑇𝑦𝑦 + 𝐹𝐹𝑔𝑔𝑔𝑔 = 𝑚𝑚𝑎𝑎𝑦𝑦

𝑇𝑇𝑦𝑦 = 8.85 × 103 N

𝐹𝐹𝑔𝑔𝑔𝑔 = −𝑚𝑚𝑚𝑚
= −(975 kg)(9.81 m/s 2 )
= −9.56 × 103 N

(8.85 × 103 − 9.56 × 103 ) N


𝑎𝑎𝑦𝑦 =
975 kg

∴ �𝑎𝑎𝑦𝑦 � = 0.733 m⁄s 2 , direction of downward.

(4.51)

Let the three masses be denoted, from left to right, as 𝑚𝑚1 , 𝑚𝑚2 , and 𝑚𝑚3 , as shown in the figure below.

We take the right direction to be +𝑥𝑥. We are told that the table is frictionless, so the only horizontal forces
are the applied force and the contact forces between the blocks. For example, 𝐹𝐹⃗12 denotes the force exerted by
block 1 on block 2. Since the blocks are in contact, they all have the same acceleration a, to the right. Newton’s
second law can be applied to each block separately:

𝐹𝐹⃗𝑎𝑎𝑎𝑎𝑎𝑎 + 𝐹𝐹⃗21 = 𝑚𝑚1 𝑎𝑎⃗


𝐹𝐹⃗12 + 𝐹𝐹⃗32 = 𝑚𝑚2 𝑎𝑎⃗
𝐹𝐹⃗23 = 𝑚𝑚3 𝑎𝑎⃗

Adding all three equations and using Newton’s third law, one finds

𝐹𝐹⃗𝑎𝑎𝑎𝑎𝑎𝑎 12 N
𝑎𝑎⃗ = = = 2.0 m⁄s 2 (to the right)
𝑚𝑚1 + 𝑚𝑚2 + 𝑚𝑚3 (1.0 + 2.0 + 3.0) kg

Thus, the force block 3 exerts on block 2 is:

∴ 𝐹𝐹32 = 𝐹𝐹12 − 𝑚𝑚2 𝑎𝑎 = 10 N − (2.0 kg)(2.0 m⁄s 2 ) = 6.0 N


(4.58)

Given the spring force 𝐹𝐹𝑠𝑠𝑠𝑠 and the spring constant 𝑘𝑘, the length stretched can be calculated by using Hooke’s
law in Equation 4.9: 𝐹𝐹𝑝𝑝 = −𝑘𝑘𝑘𝑘 (the negative sign means that the spring force opposes the distortion). The
spring stretches until the acceleration of both masses is the same.)

The magnitude of the spring tension is given by Hooke’s law, 𝐹𝐹𝑠𝑠𝑠𝑠 = 𝑘𝑘|𝑥𝑥|, where |𝑥𝑥| is the stretch of the
spring. The horizontal component of Newton’s second law applied to each mass gives
(1)
𝐹𝐹𝑛𝑛𝑛𝑛𝑛𝑛 = 𝐹𝐹𝑎𝑎𝑎𝑎𝑝𝑝 − 𝐹𝐹𝑠𝑠𝑠𝑠 = 𝑚𝑚3 𝑎𝑎
(2)
𝐹𝐹𝑛𝑛𝑛𝑛𝑛𝑛 = 𝐹𝐹𝑠𝑠𝑠𝑠 = 𝑚𝑚2 𝑎𝑎

as indicated in the sketch below. Adding these two equations, the acceleration of the entire system is:

𝐹𝐹𝑎𝑎𝑎𝑎𝑎𝑎 15 N
𝑎𝑎 = = = 3.0 m⁄s 2
𝑚𝑚2 + 𝑚𝑚3 (2.0 + 3.0) kg

The spring force is therefore

𝐹𝐹𝑠𝑠𝑠𝑠 = 𝑚𝑚2 𝑎𝑎 = (2.0 kg)(3.0 m⁄s 2 ) = 6.0 N

Applying Hooke’s law, the spring stretches a distance:

𝐹𝐹𝑠𝑠𝑠𝑠 6.0 N
|𝑥𝑥| = = = 0.0333 m
𝑘𝑘 180 N⁄m

∴ |𝑥𝑥| = 3.3 cm

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