Answer:
The distance the train travels before coming to a (complete) stop = 40/81 km which is approximately 493.83 meters
Explanation:
The initial speed of the train u = 80 km/h = 22 2/9 m/s = 22.
m/s
The magnitude of the constant acceleration with which the train slows, a = 0.5 m/s²
Therefore, we have the following suitable kinematic equation of motion;
v² = u² - 2 × a × s
Where;
v = The final velocity = 0 (The train comes to a stop)
s = The distance the train travels before coming to a stop
Substituting the values gives;
0² = 22.
² - 2 × 0.5 × s
2 × 0.5 × s = 22.
²
s = 22.
²/1 = 493 67/81 m = 40/81 km
The distance the train travels before coming to a (complete) stop = 40/81 km ≈ 493.83 m.
Answer:
303.29N and 1.44m/s^2
Explanation:
Make sure to label each vector with none, mg, fk, a, FN or T
Given
Mass m = 68.0 kg
Angle θ = 15.0°
g = 9.8m/s^2
Coefficient of static friction μs = 0.50
Coefficient of kinetic friction μk =0.35
Solution
Vertically
N = mg - Fsinθ
Horizontally
Fs = F cos θ
μsN = Fcos θ
μs( mg- Fsinθ) = Fcos θ
μsmg - μsFsinθ = Fcos θ
μsmg = Fcos θ + μsFsinθ
F = μsmg/ cos θ + μs sinθ
F = 0.5×68×9.8/cos 15×0.5×sin15
F = 332.2/0.9659+0.5×0.2588
F =332.2/1.0953
F = 303.29N
Fnet = F - Fk
ma = F - μkN
a = F - μk( mg - Fsinθ)
a = 303.29 - 0.35(68.0 * 9.8- 303.29*sin15)/68.0
303.29-0.35( 666.4 - 303.29*0.2588)/68.0
303.29-0.35(666.4-78.491)/68.0
303.29-0.35(587.90)/68.0
(303.29-205.45)/68.0
97.83/68.0
a = 1.438m/s^2
a = 1.44m/s^2