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Sagot :
Certainly! Let's solve this problem step-by-step.
1. Angle Conversion:
- We start by converting the angle of inclination from degrees to radians because calculations in physics often use the radian measure of angles.
- The angle given is [tex]\(30^\circ\)[/tex]. In radians, this is [tex]\(0.5236\)[/tex] radians (approximated to four decimal places).
2. Constants:
- The acceleration due to gravity ([tex]\(g\)[/tex]) is [tex]\(9.81 \, \text{m/s}^2\)[/tex].
3. Calculating the Acceleration Down the Plane:
- The acceleration [tex]\(a\)[/tex] down an inclined plane is given by [tex]\(a = g \sin(\theta)\)[/tex], where [tex]\(\theta\)[/tex] is the angle of inclination.
- Substitute [tex]\(g = 9.81 \, \text{m/s}^2\)[/tex] and [tex]\(\theta = 0.5236 \, \text{radians}\)[/tex]. This results in:
- [tex]\(a = 9.81 \times \sin(0.5236) = 4.905 \, \text{m/s}^2\)[/tex].
4. Calculating the Distance Slid:
- Using the equation of motion [tex]\(s = ut + \frac{1}{2} a t^2\)[/tex], and knowing that the initial velocity [tex]\(u\)[/tex] is zero and the time [tex]\(t\)[/tex] is [tex]\(4\)[/tex] seconds, we can calculate the distance [tex]\(s\)[/tex] slid.
- Substituting [tex]\(a = 4.905 \, \text{m/s}^2\)[/tex] and [tex]\(t = 4 \, \text{s}\)[/tex]:
- [tex]\(s = 0 \cdot 4 + \frac{1}{2} \cdot 4.905 \cdot 4^2 = 0.5 \times 4.905 \times 16 = 39.24 \, \text{m}\)[/tex].
5. Calculating the Coefficient of Kinetic Friction ([tex]\( \mu_k \)[/tex]):
- The coefficient of kinetic friction on an inclined plane is given by [tex]\( \mu_k = \tan(\theta) \)[/tex].
- Substitute [tex]\(\theta = 0.5236 \, \text{radians}\)[/tex]:
- [tex]\(\mu_k = \tan(0.5236) = 0.5774\)[/tex] (approximated to four decimal places).
6. Estimating the Coefficient of Static Friction ([tex]\(\mu_s\)[/tex]):
- Typically, the coefficient of static friction ([tex]\(\mu_s\)[/tex]) is slightly higher than the coefficient of kinetic friction ([tex]\(\mu_k\)[/tex]).
- Assuming [tex]\(\mu_s\)[/tex] is [tex]\(0.1\)[/tex] units higher than [tex]\(\mu_k\)[/tex]:
- [tex]\(\mu_s = 0.5774 + 0.1 = 0.6774\)[/tex].
By following these steps, we have determined the important quantities for the problem:
- Angle in radians: [tex]\(0.5236\)[/tex]
- Acceleration down the plane: [tex]\(4.905 \, \text{m/s}^2\)[/tex]
- Distance slid: [tex]\(39.24 \, \text{m}\)[/tex]
- Coefficient of kinetic friction ([tex]\(\mu_k\)[/tex]): [tex]\(0.5774\)[/tex]
- Coefficient of static friction ([tex]\(\mu_s\)[/tex]): [tex]\(0.6774\)[/tex].
These values should help you understand the sliding dynamics of the object on the inclined plane.
1. Angle Conversion:
- We start by converting the angle of inclination from degrees to radians because calculations in physics often use the radian measure of angles.
- The angle given is [tex]\(30^\circ\)[/tex]. In radians, this is [tex]\(0.5236\)[/tex] radians (approximated to four decimal places).
2. Constants:
- The acceleration due to gravity ([tex]\(g\)[/tex]) is [tex]\(9.81 \, \text{m/s}^2\)[/tex].
3. Calculating the Acceleration Down the Plane:
- The acceleration [tex]\(a\)[/tex] down an inclined plane is given by [tex]\(a = g \sin(\theta)\)[/tex], where [tex]\(\theta\)[/tex] is the angle of inclination.
- Substitute [tex]\(g = 9.81 \, \text{m/s}^2\)[/tex] and [tex]\(\theta = 0.5236 \, \text{radians}\)[/tex]. This results in:
- [tex]\(a = 9.81 \times \sin(0.5236) = 4.905 \, \text{m/s}^2\)[/tex].
4. Calculating the Distance Slid:
- Using the equation of motion [tex]\(s = ut + \frac{1}{2} a t^2\)[/tex], and knowing that the initial velocity [tex]\(u\)[/tex] is zero and the time [tex]\(t\)[/tex] is [tex]\(4\)[/tex] seconds, we can calculate the distance [tex]\(s\)[/tex] slid.
- Substituting [tex]\(a = 4.905 \, \text{m/s}^2\)[/tex] and [tex]\(t = 4 \, \text{s}\)[/tex]:
- [tex]\(s = 0 \cdot 4 + \frac{1}{2} \cdot 4.905 \cdot 4^2 = 0.5 \times 4.905 \times 16 = 39.24 \, \text{m}\)[/tex].
5. Calculating the Coefficient of Kinetic Friction ([tex]\( \mu_k \)[/tex]):
- The coefficient of kinetic friction on an inclined plane is given by [tex]\( \mu_k = \tan(\theta) \)[/tex].
- Substitute [tex]\(\theta = 0.5236 \, \text{radians}\)[/tex]:
- [tex]\(\mu_k = \tan(0.5236) = 0.5774\)[/tex] (approximated to four decimal places).
6. Estimating the Coefficient of Static Friction ([tex]\(\mu_s\)[/tex]):
- Typically, the coefficient of static friction ([tex]\(\mu_s\)[/tex]) is slightly higher than the coefficient of kinetic friction ([tex]\(\mu_k\)[/tex]).
- Assuming [tex]\(\mu_s\)[/tex] is [tex]\(0.1\)[/tex] units higher than [tex]\(\mu_k\)[/tex]:
- [tex]\(\mu_s = 0.5774 + 0.1 = 0.6774\)[/tex].
By following these steps, we have determined the important quantities for the problem:
- Angle in radians: [tex]\(0.5236\)[/tex]
- Acceleration down the plane: [tex]\(4.905 \, \text{m/s}^2\)[/tex]
- Distance slid: [tex]\(39.24 \, \text{m}\)[/tex]
- Coefficient of kinetic friction ([tex]\(\mu_k\)[/tex]): [tex]\(0.5774\)[/tex]
- Coefficient of static friction ([tex]\(\mu_s\)[/tex]): [tex]\(0.6774\)[/tex].
These values should help you understand the sliding dynamics of the object on the inclined plane.
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