Question

1)A 1kg box starts up a 2 degrees incline with a speed of 5m/s. How far will the box slide up the incline if the coefficient of kinetic friction between the box and incline is 0.4
2)A disk starting from rest rotates about its central axis with constant angular acceleration.In 3s, it rotates 27 rad .During that time, determine a)the angular acceleration
b)the instantaneous angular velocity of the disk at the end of the 3s

Answer 1

Question

1) A 1 kg box starts up a 2 degrees incline with a speed of $5\mathrm{m / s}$ . How far will the box slide up the incline if the coefficient of kinetic friction between the box and incline is 0.4?

2) A disc starting from rest rotates about its axis with constant angular acceleration. In 3 s, it rotates 27 rad. During that time, determine a) the angular acceleration b) the instantaneous angular velocity of the disc at the end of the 3 s.

Answer

1)

The second Newton's Law

$$O X: - m a = - F _ {r} - m g * \sin (\alpha)$$ $$O Y: N = m g * \cos (\alpha)$$ $$F _ {r} = \mu N \rightarrow a = g (\sin (\alpha) + \mu * \cos (\alpha))$$

The equation for instantaneous velocity

$$v _ {f} = 0 = v _ {0} - a t \rightarrow a = \frac {v _ {0}}{t}$$

From the last two equations we obtain

$$t = \frac {v _ {0}}{g (\sin (\alpha) + \mu * \cos (\alpha))}$$

And distance that box reached

$$\boldsymbol {x} = v _ {0} t - \frac {a t ^ {2}}{2} = \frac {v _ {0} t}{2} = \frac {v _ {0} ^ {2}}{2 g (\sin (\alpha) + \mu * \cos (\alpha))} \approx 2. 9 3 \boldsymbol {m}$$

2) a)

The equation of motion for rotational motion

$$\varphi = \varphi_ {0} + \omega_ {0} t + \frac {\beta t ^ {2}}{2}$$

In this case we obtain

$$\varphi = \frac {\beta t ^ {2}}{2} \rightarrow \boldsymbol {\beta} = \frac {2 \varphi}{t ^ {2}} = 6 \frac {\boldsymbol {r a d}}{s ^ {2}}$$

2) b)

The equation for instantaneous velocity

$$\omega_ {3} = \beta t = 1 8 \frac {\text {rad}}{\text {s}}$$

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