Question

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?

Answer 1

Answer:

a). 1.18 m

b). 0.67 s

c). 3.5 $\frac{m}{s}$

Explanation:

$a=-g*sen*(32)= -5.1932 \frac{m}{s^{2} }= -5.2 \frac{m}{s^{2} }$

a).

$V_{f} ^{2} =V_{i} ^{2}+2*a*d$ , $V_{f} =0$, $V_{i} =3.5\frac{m}{s }$

$d= \frac{3.5^{2}\frac{m}{s} }{2*5.2\frac{m}{s^{2} } } = 1.179426 m\\d=1.18m$

b).

$V_{f} = V_{i} + a *t$ , $V_{f} =0$, $V_{i} =3.5\frac{m}{s }$

$0 = 3.5 + 5.2*t$

$t= \frac{3.5 \frac{m}{s} }{5.2 \frac{m}{s^{2} } }$

$t= 0.67 s$

c).

$V_{f} ^{2} =V_{i} ^{2}+2*a*d$, $V_{i} =0$

$V_{f} ^{2} = 0 + 2* 5.2 \frac{m}{s^{2} } * 1.18m = 12.272 \frac{m^{2} }{s^{2} }$

$V_{f} =\sqrt{12.272 \frac{m^{2} }{s^{2} } } =3.503 \frac{m}{s}$