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

Answer on Question #52867-Physics-Mechanics-Kinematics-Dynamics

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?

Solution

Let us first find the acceleration of the box. Let's write all forces that act on the box:

$$m \vec {g} + \vec {N} + \overrightarrow {F _ {f r}} = m \vec {a}$$

Then projected the forces on axis $x$ and $y$ :

$$- m g \sin \theta - F _ {f r} = m a, N - m g \cos \theta = 0.$$

By the definition, the friction force is $F_{fr} = \mu_k N = \mu_k m g \cos \theta$ , and we can find the acceleration of the box from the first equation:

$$- m g \sin \theta - \mu_ {k} m g \cos \theta = m a,$$ $$a = - g (\sin \theta + \mu_ {k} \cos \theta).$$

Obviously, the box will slide up the incline before the velocity of the box becomes zero and it will stop. Then, we can find the distance $s$ that the box slides up the incline before it stops from the kinematic equation:

$$v ^ {2} = v _ {0} ^ {2} + 2 a s.$$

Because $v = 0$ we get:

$$s = - \frac {v _ {0} ^ {2}}{2 a} = \frac {v _ {0} ^ {2}}{2 g (\sin \theta + \mu_ {k} \cos \theta)} = \frac {\left(5 \frac {m}{s}\right) ^ {2}}{2 \cdot 9 . 8 \frac {m}{s ^ {2}} \cdot (\sin 2 {}^ {\circ} + 0 . 4 \cdot \cos 2 {}^ {\circ})} = 2. 9 3 m.$$

Answer: 2.93 m.

2) A disc starting from rest rotates about its central axis with constant angular acceleration. In 3s, it rotates 27rad. During that time, determine

a) the angular acceleration

b) the instantaneous angular velocity of the disk at the end of the 3s.

Solution

a) By the definition, $\theta = \frac{1}{2}\alpha t^2 + \omega_i t$, where $\theta$ is the angular displacement, $\alpha$ is the angular acceleration, $t$ is the time and $\omega_i$ is the initial angular velocity. Since, $\omega_i = 0$ (disc starting rotates from rest) we get:

$$\theta = \frac{1}{2} \alpha t^2.$$

From this formula we can find the angular acceleration:

$$\alpha = \frac{2\theta}{t^2} = \frac{2 \cdot 27 \text{ rad}}{(3s)^2} = 6 \frac{\text{rad}}{s^2}.$$

b) In order to find the instantaneous angular velocity we use the formula:

$$\omega = \alpha t = 6 \frac{\text{rad}}{s^2} \cdot 3s = 18 \frac{\text{rad}}{s}.$$

Answer: $6 \frac{\text{rad}}{s^2}$; $18 \frac{\text{rad}}{s}$.