We know, Mechanical Energy = K.E. + P.E.
As ball is at ground, P.E. would be zero. But as it is in motion, it must have some K.E. and that is:
K.E. = 1/2 mv²
K.E. = 1/2 * 1 * 2²
K.E. = 4/2
K.E. = 2 J
In short, Your Answer would be Option B
Hope this helps!
<span>Amplitude is the correct answer. I hope this helps.</span>
Answer:
The answer is "
"
Explanation:
Given:
![m_i = 2000 \ kg \\\\v_i= 4.1 \ \frac{m}{s} \\\\v_f = 3.4 \ \frac{m}{s} \\](https://tex.z-dn.net/?f=m_i%20%3D%202000%20%5C%20kg%20%5C%5C%5C%5Cv_i%3D%204.1%20%5C%20%5Cfrac%7Bm%7D%7Bs%7D%20%5C%5C%5C%5Cv_f%20%3D%203.4%20%5C%20%5Cfrac%7Bm%7D%7Bs%7D%20%5C%5C)
Using formula:
![\to m_iv_i = m_fv_f \\\\\to m_f= \frac{m_iv_i}{v_f}](https://tex.z-dn.net/?f=%5Cto%20m_iv_i%20%3D%20m_fv_f%20%5C%5C%5C%5C%5Cto%20m_f%3D%20%5Cfrac%7Bm_iv_i%7D%7Bv_f%7D)
= system initial and final linear momentum.
= system original and final linear pace.
= original weight of the car freight.
= car's maximum weight
![\boxed{m_f = 2.41 \times 10^3}](https://tex.z-dn.net/?f=%5Cboxed%7Bm_f%20%3D%202.41%20%5Ctimes%2010%5E3%7D)