Respuesta :
Answer
s = 2.592 m
Explanation:
given,
initial velocity of the block = 5.04 m/s
angle of the inclination of the slope = 30.0°
final velocity = 0 m/s
to calculate the distance traveled by the vehicle
using equation of motion
v² - u² = 2 a s
a = - g sin θ
0 - 5.04² = 2 × (- g sin θ) × s
[tex]s = \dfrac{5.04^2}{2 \times 9.8 \times sin 30^0}[/tex]
s = 2.592 m
hence, the block will move a distance of 2.592 m
The distance traveled by the block before stopping on the inclination is 2.592 m.
Given to us
frictionless inclination angle = 30°,
Initial speed of the block = 5.04 m/s,
What is the acceleration of the block?
The acceleration of the block is at an angle of 30.0°,
[tex]a=-g\ sin\theta[/tex]
Substitute the values,
[tex]a = -9.80 \times Sin30^o[/tex]
a = -4.9 m/s²
Thus, the acceleration of the block as it climbs up is -4.9 m/s².
What is the distance traveled by the block?
The distance of the block can be found using the third equation of motion,
[tex]v^2-u^2 = 2as\\\\0^2-(5.04)^2 = 2 (-4.9) \times s\\\\s = 2.592\rm\ m[/tex]
Hence, the distance traveled by the block before stopping on the inclination is 2.592 m.
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