The force that is pushing the crate up the ramp is competing with at least another two forces, those being G force= defines the attraction to the earth and Ff, the friction between the crate and the rough material of the ramp
Any Force can be defined by the weigh of a particular body and the acceleration
the kinetic friction indicative would be defined by the report between the active force and the friction force, considering G
So 5000-22%×5000=250× a
a=3900/250 m/s
Answer:
About 7.67 m/s.
Explanation:
Mechanical energy is always conserved. Hence:
Where <em>U</em> is potential energy and <em>K</em> is kinetic energy.
Let the bottom of the slide be where potential energy equals zero. As a result, the final potential energy is zero. Additionally, because the child starts from rest, the initial kinetic energy is zero. Thus:
Substitute and solve for final velocity:
In conclusion, the child's speed at the bottom of the slide is about 7.67 m/s.
Answer:
3.69 m/s
Explanation:
Forces :
mgsin Θ - mumgcosΘ = ma
g x sinΘ - mu x g x cosΘ = a
9.8 x sin 21 - 0.53 x 9.8 x cos 21 = a
a = -1.337 m/s²
so you have final velocity = 0 m/s
initial velocity = ? m/s
Given d = 5.1 m
By kinematics
vf² = vo² + 2ad
0 = vo² + 2 x -1.337*5.1
vo = 3.69 m/s
Answer:
Vf = 41.6 [m/s]
Explanation:
To solve this problem we must use the equations of kinematics.
Vf² = Vo² + (2*g*y)
where:
Vf = final velocity [m/s]
Vo = initial velocity = 0
g = gravity acceleration = 9.81 [m/s²]
y = height = 88.2 [m]
Note: The positive sign of the equation tells us that the acceleration of gravity goes in the direction of motion.
Vf² = Vo² + (2*g*y)
Vf² = 0 + (2*9.81*88.2)
Vf = (1730.48)^0.5
Vf = 41.6 [m/s]
