Here Change in Kinetic Energy
= Work Done by Friction
Therefore, substituting the
given values to the equation, we get
0.5 * m * (vFinal^2 -
vInitial^2) = µ m g * d
Therefore
0.5*( 5.90^2 - Vfinal^2 ) =
0.100*9.8*2.10
Therefore
vfinal = 5.54 m/sec
<span> </span>
Vector u :
u = 6 i - 3 j
The magnitude of vector u :
| u | =
![\sqrt{6 ^{2}+(-3) ^{2} } = \sqrt{36+9}= \\ \sqrt{45}= \sqrt{9*5}=3 \sqrt{5}](https://tex.z-dn.net/?f=%20%5Csqrt%7B6%20%5E%7B2%7D%2B%28-3%29%20%5E%7B2%7D%20%20%7D%20%3D%20%5Csqrt%7B36%2B9%7D%3D%20%5C%5C%20%20%5Csqrt%7B45%7D%3D%20%5Csqrt%7B9%2A5%7D%3D3%20%20%5Csqrt%7B5%7D%20%20%20%20%20)
Answer:
The magnitude of vector u is 3√5.