A force is applied to a block sliding along a surface (Figure 2). The magnitude of the force is 15 N, and the horizontal compone
1 answer:
Answer:
Fy = 14.3 [N]
Explanation:
To be able to solve this problem we must know that the force is a vector and has magnitude and direction, therefore it can be decomposed into the force in the X & y components:
When we have the components on the horizontal and vertical axes we must use the Pythagorean theorem.
where:
F = 15 [N]
Fx = horizontal component = 4.5 [N]
Fy = vertical component [N]
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Answer:
The initial velocity is 0.5114 m/s or 511.4 mm/s
Explanation:
Let the initial velocity be 'v'.
Given:
Mass of the ball (m) = 130 g = 0.130 kg [ 1 g = 0.001 kg]
Initial height of the ball (h₁) = 1.4 mm = 0.0014 m [ 1 mm = 0.001 m]
Final height of the ball (h₂) = 15 mm = 0.015 m
Now, from conservation of energy principle, energy can neither be created nor be destroyed but converted from one form to another.
Here, the kinetic energy of the ball is converted to gravitational potential energy of the ball after reaching the final height.
Change in kinetic energy is given as:
As it just touches the 15 mm high roof, the final velocity will be zero. So,
.
Now, the change in kinetic energy is equal to:
Change in gravitational potential energy = Final PE - Initial PE
So,
[ g = 9.8 m/s²]
Now, Change in KE = Change in PE
Therefore, the initial velocity is 0.5114 m/s or 511.4 mm/s
Answer:
θ = sin⁻¹
Explanation:
From one of the equations of motion, v² = u² + 2as.......... equation 1
Since the object thrown was moving against gravity, then the acceleration, a would change to -g and the initial velocity u would change to V₀ sin θ because the object is travelling at angle of θ to the horizontal. By inputting all these parameter into equation 1, we would arrive at:
v² = (u sin θ)² - 2gd
(u sin θ)² = 2gd
d = (u sin θ)²/2g
sin² θ = 2gd
sin θ =
θ = sin⁻¹
