(A) For the system consisting of the two blocks, the change in the kinetic energy of the system is equal to work done by gravity on the system.
(D) For the system consisting of the two blocks, the pulley and the Earth, the change in the total mechanical energy of the system is zero.
<h3 /><h3>The given parameters:</h3>
- Mass of block 1 = m1
- Mass of block 2, = m2
- Height of block 1 above the ground, = h1
- Height of block 2 above the ground = h2
The total initial mechanical energy of the two block system is calculated as follows;
![m_1gh_1 + \frac{1}{2} m_1v_1_i^2 = m_2gh_2 + \frac{1}{2} m_2v_2_i^2\\\\m_1gh_1 + 0 = m_2gh_2 + 0\\\\m_1gh_1 = m_2gh_2\\\\m_1gh_1 - m_2gh_2 = 0](https://tex.z-dn.net/?f=m_1gh_1%20%2B%20%5Cfrac%7B1%7D%7B2%7D%20m_1v_1_i%5E2%20%3D%20m_2gh_2%20%2B%20%5Cfrac%7B1%7D%7B2%7D%20m_2v_2_i%5E2%5C%5C%5C%5Cm_1gh_1%20%2B%200%20%3D%20m_2gh_2%20%2B%200%5C%5C%5C%5Cm_1gh_1%20%3D%20m_2gh_2%5C%5C%5C%5Cm_1gh_1%20-%20m_2gh_2%20%3D%200)
When the block m2 reaches the ground the block m1 attains maximum height and the total mechanical energy at this point is given as;
![m_1g(h_1 + h_2) + K.E_1 = \frac{1}{2}m_2v_{max}^2 + P.E_2\\\\m_1g(h_1 + h_2 ) -PE_2 = \frac{1}{2}m_2v_{max}^2 - K.E_1\\\\m_1g(h_1 + h_2 ) - 0= \frac{1}{2}m_2v_{max}^2 - 0\\\\m_1g(h_1 + h_2 ) = \frac{1}{2}m_2v_{max}^2\\\\W = \frac{1}{2}m_2v_{max}^2](https://tex.z-dn.net/?f=m_1g%28h_1%20%2B%20h_2%29%20%2B%20K.E_1%20%3D%20%5Cfrac%7B1%7D%7B2%7Dm_2v_%7Bmax%7D%5E2%20%2B%20P.E_2%5C%5C%5C%5Cm_1g%28h_1%20%2B%20h_2%20%29%20-PE_2%20%3D%20%5Cfrac%7B1%7D%7B2%7Dm_2v_%7Bmax%7D%5E2%20-%20K.E_1%5C%5C%5C%5Cm_1g%28h_1%20%2B%20h_2%20%29%20%20-%200%3D%20%5Cfrac%7B1%7D%7B2%7Dm_2v_%7Bmax%7D%5E2%20-%200%5C%5C%5C%5Cm_1g%28h_1%20%2B%20h_2%20%29%20%20%3D%20%5Cfrac%7B1%7D%7B2%7Dm_2v_%7Bmax%7D%5E2%5C%5C%5C%5CW%20%3D%20%5Cfrac%7B1%7D%7B2%7Dm_2v_%7Bmax%7D%5E2)
Thus, we can conclude the following before the block m2 reaches the ground;
- For the system consisting of the two blocks, the pulley and the Earth, the change in the total mechanical energy of the system is zero.
- For the system consisting of the two blocks, the change in the kinetic energy of the system is equal to work done by gravity on the system.
Learn more about conservation of mechanical energy here: brainly.com/question/332163
Answer:
Explanation:
Angular momentum ( L ) = moment of inertia x angular velocity ( I X ω )
Moment of inertia of two 480 g masses about axle = 2 x mr² = 2 x 480 x10⁻³ x( 24 x 10 ⁻ 2 )² = 0. 552960 kg m².
Angular velocity = 5 rad / s.
Angular momentum = 0.552960 x 5 = 2.765 kg m2.
The direction of angular momentum will be along axle.So vector angular
momentum makes zero degree with axle.
Answer:
The distance travel before stopping is 1.84 m
Explanation:
Given :
coefficient of kinetic friction ![\mu_{k} = 0.250](https://tex.z-dn.net/?f=%5Cmu_%7Bk%7D%20%3D%200.250)
Zak's speed ![v = 3 \frac{m}{s}](https://tex.z-dn.net/?f=v%20%3D%203%20%5Cfrac%7Bm%7D%7Bs%7D)
Gravitational acceleration
![\frac{m}{s^{2} }](https://tex.z-dn.net/?f=%5Cfrac%7Bm%7D%7Bs%5E%7B2%7D%20%7D)
Work done by frictional force is given by,
![W = \Delta K](https://tex.z-dn.net/?f=W%20%3D%20%5CDelta%20K)
![\mu _{k} mg d = \frac{1}{2} m v^{2}](https://tex.z-dn.net/?f=%5Cmu%20_%7Bk%7D%20mg%20d%20%3D%20%5Cfrac%7B1%7D%7B2%7D%20m%20v%5E%7B2%7D)
![d = \frac{v^{2} }{2 g \mu _{k} }](https://tex.z-dn.net/?f=d%20%3D%20%5Cfrac%7Bv%5E%7B2%7D%20%20%7D%7B2%20g%20%5Cmu%20_%7Bk%7D%20%7D)
![d = \frac{9}{2 \times 9.8 \times 0.250}](https://tex.z-dn.net/?f=d%20%3D%20%5Cfrac%7B9%7D%7B2%20%5Ctimes%209.8%20%5Ctimes%200.250%7D)
m
Therefore, the distance travel before stopping is 1.84 m
Probably 90 j but im not sure I haven’t done any work like this in a while
ANSWER: 1.6N (forth option)
EXPLANATION:
E= Force x displacement
69=F(42)
69/42 = F
F=1.64 N
Answer : 1.6N