A 1.23-kg box slides up an incline plane with a speed of 8.6 m/s.If the angle of the plane with respect to the horizontal is 11 degrees and the coefficient of friction is 0.33,how long will it take for the box to come to rest?
A)1.7 s
B)2.5 s
C)2.7 s
D)17 s

Question

Quizplus · Accepted Answer

To find the time it takes for the box to come to rest, we first need to calculate the deceleration (a) of the box. The forces acting on the box include the component of gravity parallel to the incline, $mg\sin(\theta)$, and the frictional force, $f = \mu mg\cos(\theta)$, where $m$ is the mass of the box, $g$ is the acceleration due to gravity (9.8 m/s$^2$), $\mu$ is the coefficient of friction, and $\theta$ is the angle of the incline.The net force acting on the box along the incline is the sum of the gravitational component and the frictional force, both acting to decelerate the box. Thus, the net force $F_{\text{net}} = mg\sin(\theta) + \mu mg\cos(\theta)$.The acceleration $a$ can be found using Newton's second law, $F = ma$, so $a = \frac{F_{\text{net}}}{m}$.Substituting the given values:$$a = \frac{g(\sin(11^\circ) + \mu\cos(11^\circ))}{1} = 9.8(\sin(11^\circ) + 0.33\cos(11^\circ))$$To find the time $t$ it takes for the box to come to rest, we use the equation $v = u + at$, where $v$ is the final velocity (0 m/s, since the box comes to rest), $u$ is the initial velocity (8.6 m/s), and $a$ is the deceleration we just calculated. Rearranging for $t$, we get $t = \frac{-u}{a}$.Plugging in the values and solving for $t$ gives us the time it takes for the box to come to rest. The calculation involves trigonometric functions for the angle and accounts for both the gravitational component down the incline and the opposing frictional force. The exact numerical calculation is omitted here, but following these steps with the given values leads to the correct answer, which is approximately 1.7 seconds, making option A the correct choice.

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