A block is projected up a frictionless inclined plane with initial speed v0 = 3.50 m/s. The angle of incline is θ =32.0°. (a) How far up the plane does the block go? (b) How long does it take to get there? (c) What is its speedwhen it gets back to the bottom?

Respuesta :

Answer:

a). 1.18 m

b). 0.67 s

c). 3.5 [tex]\frac{m}{s}[/tex]

Explanation:

[tex]a=-g*sen*(32)= -5.1932 \frac{m}{s^{2} }= -5.2 \frac{m}{s^{2} }[/tex]

a).

[tex]V_{f} ^{2} =V_{i} ^{2}+2*a*d[/tex] , [tex]V_{f} =0[/tex], [tex]V_{i} =3.5\frac{m}{s } [/tex]

[tex]d= \frac{3.5^{2}\frac{m}{s}  }{2*5.2\frac{m}{s^{2} } } = 1.179426 m\\d=1.18m[/tex]

b).

[tex]V_{f} = V_{i} + a *t[/tex] , [tex]V_{f} =0[/tex], [tex]V_{i} =3.5\frac{m}{s } [/tex]

[tex]0 = 3.5 + 5.2*t[/tex]

[tex]t= \frac{3.5 \frac{m}{s} }{5.2 \frac{m}{s^{2} } }[/tex]

[tex]t= 0.67 s[/tex]

c).

[tex]V_{f} ^{2} =V_{i} ^{2}+2*a*d[/tex], [tex]V_{i} =0[/tex]

[tex]V_{f} ^{2} = 0 + 2* 5.2 \frac{m}{s^{2} } * 1.18m = 12.272 \frac{m^{2} }{s^{2} }[/tex]

[tex]V_{f} =\sqrt{12.272 \frac{m^{2} }{s^{2} } } =3.503 \frac{m}{s}[/tex]