Answer to Question #85209 in Mechanics | Relativity for Maria

Question

A crate is given an initial speed of 3.4 m/s up the 28.5 ∘ plane shown in (Figure 1). Assume μk = 0.12.
Part A
How far up the plane will it go?
Part B
How much time elapses before it returns to its starting point?

Answer 1

A) Let's first find the acceleration of the crate. Applying the Newton’s Second Law of motion in projections on $x$- and $y$-axis, we get:

$$\sum F_x = ma_x,$$ $$\sum F_y = ma_y = 0,$$ $$-mgsin \theta - F_{fr} = ma, (1)$$ $$N - mgcos \theta = 0. (2)$$

By the definition of the force of friction we have:

$$F_{fr} = \mu_{k} N,$$

here, $\mu_{k}$ is the coefficient of kinetic friction and $N$ is the normal force.

We can find the normal force from the equation (2):

$$N = mgcos \theta.$$

Then, the force of friction will be equal:

$$F_{fr} = \mu_{k} N = \mu_{k} mgcos \theta.$$

Then, substituting the force of friction into the equation (1), we can find the acceleration of the crate:

$$-mgsin \theta - \mu_{k} mgcos \theta = ma,$$ $$-mg(sin \theta + \mu_{k} cos \theta ) = ma,$$ $$a = -g(sin \theta + \mu_{k} cos \theta).$$

Let's substitute the numbers:

$$a = -9.8 \dfrac{m}{s^2}(sin28.5^{\circ} + 0.12 \cdot cos28.5^{\circ}) = -5.71 \dfrac{m}{s^2}.$$

The sign minus indicates that the crate decelerates.

We can find how far up the plane will the crate move from the kinematic equation:

$$v_f^2 = v_i^2 + 2as,$$

here, $v_i = 3.4 m/s$ is the initial speed of the crate, $v_f = 0$ is the final speed of the crate, $a$ is the acceleration of the crate and $s$ is the distance traveled by the crate from the starting point to the top of the plane where it stops.

Then, we get:

$$s = \dfrac{v_f^2 - v_i^2 }{2a} = \dfrac{0 - (3.4 \dfrac{m}{s})^2}{2 \cdot (-5.71 \dfrac{m}{s^2})} = 1.01 m.$$

B) Let’s first find the time that the crate needs to reach the top of the plane from the kinematic equation:

$$v_f = v_i + at_{up},$$ $$t_{up} = - \dfrac{v_i}{a} = - \dfrac{3.4 \dfrac{m}{s}}{-5.71 \dfrac{m}{s^2}} = 0.59 s.$$

Then, let's find the acceleration of the crate when it slides down the plane:

$$mgsin \theta - \mu_{k} mgcos \theta = ma,$$ $$a = g(sin \theta - \mu_{k} cos \theta).$$

Let's substitute the numbers:

$$a = 9.8 \dfrac{m}{s^2}(sin28.5^{\circ} - 0.12 \cdot cos28.5^{\circ}) = 3.64 \dfrac{m}{s^2}.$$

Then, we can find the time that the crate needs to reach the bottom of the plane (its starting point)

from the kinematic equation:

$$s = v_i t + \dfrac {1}{2}at_{bottom}^2,$$ $$t_{bottom} = \sqrt{\dfrac{2s}{a}} = \sqrt{\dfrac{2 \cdot 1.01 m}{3.64 \dfrac{m}{s^2}}} = 0.74 s.$$

Then, the total time that the crate needs to travel up the plane and returns to its starting point:

$$t_{tot} = t_{up} + t_{bottom} = 0.59 s + 0.74 s = 1.33 s.$$

Answer:

A) $s = 1.01 m.$

B) $t_{tot} = 1.33 s.$