Respuesta :
Answer:
a) The block goes 7.4 m up the plane
b) It takes 1.62 seconds to get there
c) The speed at the bottom is -9.1 m/s
Explanations:The initial speed, v₀ = 9.12 m/s
Angle of inclination, θ = 35°
Note that the summation forces on the block as it goes up the inclined plane is 0. That is, ΣF = 0
mgsinθ + ma = 0
mgsinθ = -ma
a = -gsinθ
The acceleration of the block up the plane is:
a = -9.8 sin 35
a = -5.62 m/s²
a) How far up the plane does it go?
Using the equation v² = v₀² + 2as
At the maximum distance up the plane, v = 0
0² = 9.12² + 2(-5.62)s
0 = 83.17 - 11.24s
11.24s = 83.17
s = 83.17/11.24
s = 7.4 m
The block goes 7.4 m up the plane
b) How long does it take to get there?
Using the equation v = v₀ + at
Since we want to calculate the time taken to reach the maximum distance, the speed at this point is 0. v = 0m/s
Substitute a = -5.62, v₀ = 9.12, and v = 0 into v = v₀ + at
0 = 9.12 + (-5.62)t
5.62t = 9.12
t = 9.12 / 5.62
t = 1.62 seconds
c) What is its speed when it gets back to the bottom?
The total time it takes the block to go up and down the plane = 2(1.62)
The total time it takes the block to go up and down the plane = 3.24 s
To calculate the speed of the block when it gets back to the bottom, use the equation v = v₀ + at
v = 9.12 + (-5.62)(3.24)
v = 9.12 - 18.21
v = -9.1 m/s
