Question

A block is placed on a long and wide inclined plane that makes angle $\theta = 45^\circ$ with the horizontal. The coefficient of kinetic friction between the block and the plane is $\mu_k = 2$. After a quick push, the block acquires velocity of $6\sqrt{2}$ m/s that makes angle $\alpha = 60^\circ$ with the line of fastest descent. The time interval during which the block is in motion is $t_0$ s. Find the value of $98\,t_0$. (take $g = 10$ m/s$^2$)

Answer 1

117.6

Answer 2

Let the inclined plane make an angle $\theta = 45^\circ$ with the horizontal. The coefficient of kinetic friction is $\mu_k = 2$. The initial velocity of the block is $v_0 = 6\sqrt{2}$ m/s at an angle $\alpha = 60^\circ$ with the line of fastest descent (down the incline). We take $g = 10$ m/s$^2$.

Let's set up a coordinate system on the inclined plane:

• $\hat{i}$ direction is along the line of fastest descent (down the incline).
• $\hat{j}$ direction is perpendicular to the line of fastest descent, across the incline.

The components of the initial velocity are: $v_{0i} = v_0 \cos\alpha = (6\sqrt{2}) \cos 60^\circ = 6\sqrt{2} \times \frac{1}{2} = 3\sqrt{2}$ m/s. $v_{0j} = v_0 \sin\alpha = (6\sqrt{2}) \sin 60^\circ = 6\sqrt{2} \times \frac{\sqrt{3}}{2} = 3\sqrt{6}$ m/s.

The forces acting on the block parallel to the inclined plane are:

1. Gravitational force component down the incline: $F_{gi} = mg \sin\theta$.
2. Kinetic friction force: $\vec{f}_k$, which opposes the velocity $\vec{v}$. Its magnitude is $f_k = \mu_k N$, where $N$ is the normal force.

From the equilibrium in the direction perpendicular to the plane: $N - mg \cos\theta = 0 \implies N = mg \cos\theta$. So, $f_k = \mu_k mg \cos\theta$.

The acceleration components are: $a_i = \frac{dv_i}{dt} = g \sin\theta - \mu_k g \cos\theta \frac{v_i}{|\vec{v}|}$ $a_j = \frac{dv_j}{dt} = - \mu_k g \cos\theta \frac{v_j}{|\vec{v}|}$

Let's calculate the constants: $g = 10$ m/s$^2$ $\theta = 45^\circ \implies \sin\theta = \cos\theta = \frac{1}{\sqrt{2}}$ $\mu_k = 2$

$g \sin\theta = 10 \times \frac{1}{\sqrt{2}} = 5\sqrt{2}$ m/s$^2$. $\mu_k g \cos\theta = 2 \times 10 \times \frac{1}{\sqrt{2}} = 10\sqrt{2}$ m/s$^2$.

So, the acceleration components are: $a_i = 5\sqrt{2} - 10\sqrt{2} \frac{v_i}{|\vec{v}|}$ $a_j = -10\sqrt{2} \frac{v_j}{|\vec{v}|}$

Since $v_{0j} = 3\sqrt{6} > 0$ and $a_j$ is always negative when $v_j > 0$, the component $v_j$ will decrease. The block stops when $\vec{v} = \vec{0}$.

Let's analyze the motion. If the velocity is purely in the $\hat{i}$ direction ($v_j=0$), then $|\vec{v}| = v_i$. In this case, $a_i = 5\sqrt{2} - 10\sqrt{2} \frac{v_i}{v_i} = 5\sqrt{2} - 10\sqrt{2} = -5\sqrt{2}$ m/s$^2$. And $a_j = -10\sqrt{2} \frac{0}{v_i} = 0$.

Consider the initial velocity components. $v_{0i} = 3\sqrt{2}$ and $v_{0j} = 3\sqrt{6}$. The initial velocity vector is $\vec{v}_0 = 3\sqrt{2} \hat{i} + 3\sqrt{6} \hat{j}$. The initial angle with the $\hat{i}$ direction is $\alpha = 60^\circ$.

Let's assume the velocity direction changes such that the friction force always has a significant opposing component. If the motion were along the $\hat{i}$ direction only, the deceleration would be $5\sqrt{2}$ m/s$^2$. The time to stop from $v_{0i} = 3\sqrt{2}$ m/s with this deceleration would be $t = \frac{0 - v_{0i}}{-5\sqrt{2}} = \frac{-3\sqrt{2}}{-5\sqrt{2}} = \frac{3}{5} = 0.6$ s.

However, $v_{0j} = 3\sqrt{6} \neq 0$. Let's consider the case where the velocity component $v_j$ becomes zero. The equation for $a_j$ is $\frac{dv_j}{dt} = -10\sqrt{2} \frac{v_j}{v}$. If we assume that the angle $\alpha$ remains constant at $60^\circ$ for some time, then $v_i/v = \cos 60^\circ = 1/2$ and $v_j/v = \sin 60^\circ = \sqrt{3}/2$. In this situation, $a_i = 5\sqrt{2} - 10\sqrt{2} \times \frac{1}{2} = 5\sqrt{2} - 5\sqrt{2} = 0$. And $a_j = -10\sqrt{2} \times \frac{\sqrt{3}}{2} = -5\sqrt{6}$ m/s$^2$.

If $a_i = 0$, then $v_i(t) = v_{0i} = 3\sqrt{2}$ m/s (constant). If $a_j = -5\sqrt{6}$ m/s$^2$, then $v_j(t) = v_{0j} + a_j t = 3\sqrt{6} - 5\sqrt{6} t$.

The block will stop when $v_j$ becomes zero (assuming $v_i$ is still positive). $v_j(t) = 0 \implies 3\sqrt{6} - 5\sqrt{6} t = 0 \implies t = \frac{3\sqrt{6}}{5\sqrt{6}} = \frac{3}{5} = 0.6$ s.

At $t = 0.6$ s, the velocity is $\vec{v}(0.6) = v_i(0.6) \hat{i} + v_j(0.6) \hat{j} = 3\sqrt{2} \hat{i} + 0 \hat{j}$. So, at $t = 0.6$ s, the velocity is purely in the $\hat{i}$ direction.

Now, the acceleration components are: $a_i = 5\sqrt{2} - 10\sqrt{2} \frac{v_i}{|\vec{v}|} = 5\sqrt{2} - 10\sqrt{2} \frac{3\sqrt{2}}{3\sqrt{2}} = 5\sqrt{2} - 10\sqrt{2} = -5\sqrt{2}$ m/s$^2$. $a_j = -10\sqrt{2} \frac{v_j}{|\vec{v}|} = -10\sqrt{2} \frac{0}{3\sqrt{2}} = 0$.

From $t = 0.6$ s onwards, the block moves with constant deceleration $a_i = -5\sqrt{2}$ m/s$^2$ along the $\hat{i}$ direction. The velocity at $t = 0.6$ s is $v_i(0.6) = 3\sqrt{2}$ m/s. The time taken to stop from this point is $t' = \frac{0 - v_i(0.6)}{a_i} = \frac{-3\sqrt{2}}{-5\sqrt{2}} = \frac{3}{5} = 0.6$ s.

The total time the block is in motion is $t_0 = 0.6 \text{ s} + 0.6 \text{ s} = 1.2$ s.

The question asks for the value of $98\,t_0$. $98\,t_0 = 98 \times 1.2 = 117.6$.