Answer :
To find the final velocity of the truck sliding down a frictionless hill, we will use principles from physics, particularly the concepts of kinematics and gravitational acceleration along an incline.
Here's how we can break down the problem step-by-step:
1. Identify the given parameters:
- Mass of the truck, [tex]\( m = 3220 \, \text{kg} \)[/tex]
- Initial velocity, [tex]\( v_i = 4.71 \, \text{m/s} \)[/tex]
- Distance along the slope, [tex]\( d = 22.0 \, \text{m} \)[/tex]
- Slope angle, [tex]\( \theta = 8.30^\circ \)[/tex]
2. Convert the slope angle to radians:
The slope angle in radians is:
[tex]\[ \theta_{\text{radians}} = \theta \times \frac{\pi}{180} \][/tex]
[tex]\[ \theta_{\text{radians}} = 8.30 \times \frac{\pi}{180} \approx 0.1448 \, \text{radians} \][/tex]
3. Calculate the gravitational acceleration component along the slope:
The gravitational force causes an acceleration down the slope, which is given by:
[tex]\[ a_{\parallel} = g \sin(\theta) \][/tex]
where [tex]\( g = 9.81 \, \text{m/s}^2 \)[/tex]. So,
[tex]\[ a_{\parallel} = 9.81 \times \sin(0.1448) \approx 1.414 \, \text{m/s}^2 \][/tex]
4. Use the kinematic equation to find the final velocity:
The kinematic equation relating initial velocity [tex]\( v_i \)[/tex], final velocity [tex]\( v_f \)[/tex], acceleration [tex]\( a \)[/tex], and distance [tex]\( d \)[/tex] is:
[tex]\[ v_f^2 = v_i^2 + 2a_{\parallel} d \][/tex]
Plug in the known values:
[tex]\[ v_f^2 = (4.71)^2 + 2 \times 1.414 \times 22.0 \][/tex]
[tex]\[ v_f^2 = 22.1841 + 62.328 \][/tex]
[tex]\[ v_f^2 = 84.5121 \][/tex]
5. Solve for [tex]\( v_f \)[/tex]:
Take the square root of both sides to find the final velocity [tex]\( v_f \)[/tex]:
[tex]\[ v_f = \sqrt{84.5121} \approx 9.192 \, \text{m/s} \][/tex]
Therefore, the final velocity of the truck after sliding down the 22.0 meter frictionless hill is approximately:
[tex]\[ \boxed{9.19 \, \text{m/s}} \][/tex]
Here's how we can break down the problem step-by-step:
1. Identify the given parameters:
- Mass of the truck, [tex]\( m = 3220 \, \text{kg} \)[/tex]
- Initial velocity, [tex]\( v_i = 4.71 \, \text{m/s} \)[/tex]
- Distance along the slope, [tex]\( d = 22.0 \, \text{m} \)[/tex]
- Slope angle, [tex]\( \theta = 8.30^\circ \)[/tex]
2. Convert the slope angle to radians:
The slope angle in radians is:
[tex]\[ \theta_{\text{radians}} = \theta \times \frac{\pi}{180} \][/tex]
[tex]\[ \theta_{\text{radians}} = 8.30 \times \frac{\pi}{180} \approx 0.1448 \, \text{radians} \][/tex]
3. Calculate the gravitational acceleration component along the slope:
The gravitational force causes an acceleration down the slope, which is given by:
[tex]\[ a_{\parallel} = g \sin(\theta) \][/tex]
where [tex]\( g = 9.81 \, \text{m/s}^2 \)[/tex]. So,
[tex]\[ a_{\parallel} = 9.81 \times \sin(0.1448) \approx 1.414 \, \text{m/s}^2 \][/tex]
4. Use the kinematic equation to find the final velocity:
The kinematic equation relating initial velocity [tex]\( v_i \)[/tex], final velocity [tex]\( v_f \)[/tex], acceleration [tex]\( a \)[/tex], and distance [tex]\( d \)[/tex] is:
[tex]\[ v_f^2 = v_i^2 + 2a_{\parallel} d \][/tex]
Plug in the known values:
[tex]\[ v_f^2 = (4.71)^2 + 2 \times 1.414 \times 22.0 \][/tex]
[tex]\[ v_f^2 = 22.1841 + 62.328 \][/tex]
[tex]\[ v_f^2 = 84.5121 \][/tex]
5. Solve for [tex]\( v_f \)[/tex]:
Take the square root of both sides to find the final velocity [tex]\( v_f \)[/tex]:
[tex]\[ v_f = \sqrt{84.5121} \approx 9.192 \, \text{m/s} \][/tex]
Therefore, the final velocity of the truck after sliding down the 22.0 meter frictionless hill is approximately:
[tex]\[ \boxed{9.19 \, \text{m/s}} \][/tex]
