<u>Given</u><u>:</u>
- An object has a forward force = 100N
- An object has a reverse force = 25N
<u>To</u><u> </u><u>find</u><u> </u><u>out</u><u>:</u>
What is the resultant force?
<u>Solution</u><u>:</u>
Resultant Force = Forward force + Reserve force
= 100 N + ( - 25 N )
= 75 N
Answer:
Explanation:
Given that,
Mass of the heavier car m_1 = 1750 kg
Mass of the lighter car m_2 = 1350 kg
The speed of the lighter car just after collision can be represented as follows
![m_1u_1+m_2u_2=m_1v_1+m_2v_2\\\\v_2=\frac{m_1u_1+m_2u_2-m_1v_1}{m_2}](https://tex.z-dn.net/?f=m_1u_1%2Bm_2u_2%3Dm_1v_1%2Bm_2v_2%5C%5C%5C%5Cv_2%3D%5Cfrac%7Bm_1u_1%2Bm_2u_2-m_1v_1%7D%7Bm_2%7D)
![v_2=\frac{(1850)(1.4)+(1450)(-1.10)-(1850)(0.250)}{1450} \\\\=\frac{2590+(-1595)-(462.5)}{1450} \\\\=\frac{2590-1595-462.5}{1450} \\\\=\frac{532.5}{1450}\\\\=0.367m/s](https://tex.z-dn.net/?f=v_2%3D%5Cfrac%7B%281850%29%281.4%29%2B%281450%29%28-1.10%29-%281850%29%280.250%29%7D%7B1450%7D%20%5C%5C%5C%5C%3D%5Cfrac%7B2590%2B%28-1595%29-%28462.5%29%7D%7B1450%7D%20%5C%5C%5C%5C%3D%5Cfrac%7B2590-1595-462.5%7D%7B1450%7D%20%5C%5C%5C%5C%3D%5Cfrac%7B532.5%7D%7B1450%7D%5C%5C%5C%5C%3D0.367m%2Fs)
b) the change in the combined kinetic energy of the two-car system during this collision
![\Delta K.E=(\frac{1}{2} m_1v_1^2+\frac{1}{2} m_2v_2^2)-(\frac{1}{2} m_1u_1^2+\frac{1}{2} m_2u_2^2)\\\\=\frac{1}{2} (m_1(v_1^2-u_1^2)+m_2(v_2^2-u_2^2))](https://tex.z-dn.net/?f=%5CDelta%20K.E%3D%28%5Cfrac%7B1%7D%7B2%7D%20m_1v_1%5E2%2B%5Cfrac%7B1%7D%7B2%7D%20m_2v_2%5E2%29-%28%5Cfrac%7B1%7D%7B2%7D%20m_1u_1%5E2%2B%5Cfrac%7B1%7D%7B2%7D%20m_2u_2%5E2%29%5C%5C%5C%5C%3D%5Cfrac%7B1%7D%7B2%7D%20%28m_1%28v_1%5E2-u_1%5E2%29%2Bm_2%28v_2%5E2-u_2%5E2%29%29)
substitute the value in the equation above
![=\frac{1}{2} (1850((0.250)^2-(1.4)^2)+(1450((0.3670)^2-(-1.10)^2)\\\\=\frac{1}{2}(11850(0.0625-1.96)+(1450(0.1347)-(1.21))\\\\= \frac{1}{2}(11850(-1.8975))+(1450(-1.0753))\\\\=\frac{1}{2} (-3510.375+(-1559.185)\\\\=\frac{1}{2} (-5069.56)\\\\=-2534.78J](https://tex.z-dn.net/?f=%3D%5Cfrac%7B1%7D%7B2%7D%20%281850%28%280.250%29%5E2-%281.4%29%5E2%29%2B%281450%28%280.3670%29%5E2-%28-1.10%29%5E2%29%5C%5C%5C%5C%3D%5Cfrac%7B1%7D%7B2%7D%2811850%280.0625-1.96%29%2B%281450%280.1347%29-%281.21%29%29%5C%5C%5C%5C%3D%20%5Cfrac%7B1%7D%7B2%7D%2811850%28-1.8975%29%29%2B%281450%28-1.0753%29%29%5C%5C%5C%5C%3D%5Cfrac%7B1%7D%7B2%7D%20%28-3510.375%2B%28-1559.185%29%5C%5C%5C%5C%3D%5Cfrac%7B1%7D%7B2%7D%20%28-5069.56%29%5C%5C%5C%5C%3D-2534.78J)
Hence, the change in combine kinetic energy is -2534.78J
Answer:
E = 0.01 J
Explanation:
Given that,
The mass of the cart, m = 0.15 kg
The force constant of the spring, k = 3.58 N/m
The amplitude of the oscillations, A = 7.5 cm = 0.075 m
We need to find the total mechanical energy of the system. It can be given by the formula as follows :
![E=\dfrac{1}{2}kA^2](https://tex.z-dn.net/?f=E%3D%5Cdfrac%7B1%7D%7B2%7DkA%5E2)
Put all the values,
![E=\dfrac{1}{2}\times 3.58\times (0.075)^2\\\\=0.01\ J](https://tex.z-dn.net/?f=E%3D%5Cdfrac%7B1%7D%7B2%7D%5Ctimes%203.58%5Ctimes%20%280.075%29%5E2%5C%5C%5C%5C%3D0.01%5C%20J)
So, the value of total mechanical energy is equal to 0.01 J.
Answer:
W = 0J
Explanation:
The work done by the dresser is described as
W = f d (cos θ)
F has been given as the weight of this dresser. And it is 3500 N
d = 0 m
When you put these values into the equation
W = 3500 x 0 x cosθ
W = 0 J
This value tells us that the work done on this dresser is zero. No work has been done. Therefore the last option answers the question.