A plank with a box on it at one end is gradually raised about the other end. As the angle of inclination with the horizontal reaches 30$^{\circ}$, the box starts to slip and slides 4.0 m down the plank in 4.0 s The coefficients of static and kinetic friction between the box and the plank will be, respectively

Question

A rocket of initial mass 6000 kg ejects gases at a constant rate of 16 kg/s with a constant relative speed of 11 km/s. What is the acceleration of the rocket one minute after the blast?

Answer 1

To solve this problem, we need to determine the coefficients of static and kinetic friction between the box and the plank. The scenario provided describes a box starting to slip at a 30° angle of inclination and sliding 4.0 meters down the plank in 4.0 seconds.

First, calculate the coefficient of static friction $ \mu_s $ using the angle of inclination. The box begins to slip at a 30° angle; hence, this is when static friction is overcome:
$\mu_s = \tan \theta$
Substituting the given angle:
$\mu_s = \tan 30^\circ = \frac{1}{\sqrt{3}} \approx 0.577$
For practical purposes, we can approximate this value to 0.6, which matches one of the provided options.
Next, calculate the coefficient of kinetic friction $ \mu_k $ using the motion of the box. Use the kinematic equations to determine the acceleration of the box:
s = ut + \frac{1}{2}at^2
- Initial velocity ($u$) is 0 (starting from rest). - Displacement ($s$) is 4.0 meters, and time ($t$) is 4.0 seconds. Plugging in known values:
4.0 = 0 + \frac{1}{2} a (4.0)^2

4.0 = 8a

a = 0.5 \, \text{m/s}^2
Now, consider the forces acting on the box: gravity and friction:
a = g (\sin \theta - \mu_k \cos \theta)
- Known: $g = 9.8 \, \text{m/s}^2$, $ \sin 30^\circ = 0.5$, $ \cos 30^\circ = \frac{\sqrt{3}}{2} \approx 0.866$ Substituting the known values:
0.5 = 9.8 \times (0.5 - \mu_k \times 0.866)

0.5 = 4.9 - 9.8 \mu_k \times 0.866

9.8 \mu_k \times 0.866 = 4.4

$\mu_k \approx 0.5$
Based on these calculations, the coefficient of static friction is approximately 0.6, and the coefficient of kinetic friction is 0.5.

Thus, the correct answer is that the coefficients of static and kinetic friction are 0.6 and 0.5, respectively.

Answer 2

To solve this problem, we need to determine the coefficients of static and kinetic friction between the box and the plank. The scenario provided describes a box starting to slip at a 30° angle of inclination and sliding 4.0 meters down the plank in 4.0 seconds.

First, calculate the coefficient of static friction $ \mu_s $ using the angle of inclination. The box begins to slip at a 30° angle; hence, this is when static friction is overcome:
$\mu_s = \tan \theta$
Substituting the given angle:
$\mu_s = \tan 30^\circ = \frac{1}{\sqrt{3}} \approx 0.577$
For practical purposes, we can approximate this value to 0.6, which matches one of the provided options.
Next, calculate the coefficient of kinetic friction $ \mu_k $ using the motion of the box. Use the kinematic equations to determine the acceleration of the box:
s = ut + \frac{1}{2}at^2
- Initial velocity ($u$) is 0 (starting from rest). - Displacement ($s$) is 4.0 meters, and time ($t$) is 4.0 seconds. Plugging in known values:
4.0 = 0 + \frac{1}{2} a (4.0)^2

4.0 = 8a

a = 0.5 \, \text{m/s}^2
Now, consider the forces acting on the box: gravity and friction:
a = g (\sin \theta - \mu_k \cos \theta)
- Known: $g = 9.8 \, \text{m/s}^2$, $ \sin 30^\circ = 0.5$, $ \cos 30^\circ = \frac{\sqrt{3}}{2} \approx 0.866$ Substituting the known values:
0.5 = 9.8 \times (0.5 - \mu_k \times 0.866)

0.5 = 4.9 - 9.8 \mu_k \times 0.866

9.8 \mu_k \times 0.866 = 4.4

$\mu_k \approx 0.5$
Based on these calculations, the coefficient of static friction is approximately 0.6, and the coefficient of kinetic friction is 0.5.

Thus, the correct answer is that the coefficients of static and kinetic friction are 0.6 and 0.5, respectively.

The Correct Option is D

Solution and Explanation

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