Answer:
Energia cinetica=187.5 [J]
Explanation:
La energia cinetica se define por medio de la siguiente ecuacion:
![E_{c} =\frac{1}{2} *m*v^{2} \\donde:\\m =masa = 600[g] = 0.6[kg]\\v=velocidad= 90 [km/h]](https://tex.z-dn.net/?f=E_%7Bc%7D%20%3D%5Cfrac%7B1%7D%7B2%7D%20%2Am%2Av%5E%7B2%7D%20%5C%5Cdonde%3A%5C%5Cm%20%3Dmasa%20%3D%20600%5Bg%5D%20%3D%200.6%5Bkg%5D%5C%5Cv%3Dvelocidad%3D%2090%20%5Bkm%2Fh%5D)
Debemos convertir las unidades de kilómetros por hora a metros por segundo
![90\frac{km}{h} *\frac{1h}{3600s}*\frac{1000m}{1km}=25[m/s]](https://tex.z-dn.net/?f=90%5Cfrac%7Bkm%7D%7Bh%7D%20%2A%5Cfrac%7B1h%7D%7B3600s%7D%2A%5Cfrac%7B1000m%7D%7B1km%7D%3D25%5Bm%2Fs%5D)
Reemplazamos en la ecuacion de energia cinetica.
![E_{k} =\frac{1}{2} *0.6*(25)^{2} \\E_{k} =187.5[J]](https://tex.z-dn.net/?f=E_%7Bk%7D%20%3D%5Cfrac%7B1%7D%7B2%7D%20%2A0.6%2A%2825%29%5E%7B2%7D%20%5C%5CE_%7Bk%7D%20%3D187.5%5BJ%5D)
Can we see the diagram? Thanks.
Answer:
3.2N
Explanation:
Given parameters:
Mass of block = 1.5kg
Coefficient of kinetic friction = 0.6
Force of pull on block = 12N
Unknown:
Net force on the block = ?
Solution:
Frictional force is a force that opposes motion:
Net force = Force of pull - Frictional force
Frictional force = umg
u is coefficient of kinetic friction
m is the mass
g is the acceleration due to gravity
Frictional force = 0.6 x 1.5 x 9.8 = 8.8N
Net force = 12N - 8.8N = 3.2N