A 5 kg block is pushed up a 37 degree incline given an initial
velocity of 2 m/s at the base of a frictionless plane. The maximum
vertical displacement of the block is nearly what?
Figure the initial KE, then the height where PE is equal to that.
To find the maximum vertical displacement of the block, we can break down the motion into its horizontal and vertical components.
First, let's analyze the horizontal motion. We can use the initial velocity of 2 m/s to find the horizontal component of velocity using trigonometry. The horizontal component (𝑣𝑥) can be calculated as follows:
𝑣𝑥 = 𝑣 × cos(𝜃)
where 𝑣 is the initial velocity and 𝜃 is the incline angle.
Substituting the given values:
𝑣 = 2 m/s and 𝜃 = 37 degrees
𝑣𝑥 = 2 × cos(37) = 2 × 0.7986 = 1.597 m/s
Now let's analyze the vertical motion. We can calculate the vertical displacement (𝑦) using the kinematic equation:
𝑦 = 𝑣𝑦𝑜𝑡 × 𝑡 + (1/2) × 𝑎 × 𝑡²
where 𝑣𝑦𝑜𝑡 is the initial vertical component of velocity, 𝑡 is the time, and 𝑎 is the acceleration.
Since the block is moving up the incline, 𝑣𝑦𝑜𝑡 is positive. The initial vertical component of velocity (𝑣𝑦𝑜𝑡) can be calculated using trigonometry. It is given by:
𝑣𝑦𝑜𝑡 = 𝑣 × sin(𝜃)
Substituting the given values:
𝑣 = 2 m/s and 𝜃 = 37 degrees
𝑣𝑦𝑜𝑡 = 2 × sin(37) = 2 × 0.6018 = 1.204 m/s
Since the incline is frictionless, there is no horizontal acceleration. Therefore, 𝑎 = 0.
Using the kinematic equation, we can now calculate the vertical displacement (𝑦) at the peak of the incline. At the peak, the vertical component of velocity (𝑣𝑦) becomes 0.
0 = 𝑣𝑦𝑜𝑡 × 𝑡 + (1/2) × 𝑎 × 𝑡²
Plugging in the values:
0 = 1.204 × 𝑡 + (1/2) × 0 × 𝑡²
0 = 1.204 × 𝑡
Since 𝑡 ≠ 0, we can divide both sides of the equation by 𝑡:
0 = 1.204
This implies that the vertical displacement at the peak is 0, which means that the block doesn't reach any height above the starting point.