A block of 7.80 kg kept on an inclined plane just begins to slide at an angle of inclination of 35.0°. Once it has been set into
1 answer:
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Answer:
P= 454.11 N
Explanation:
Since P is the only horizontal force acting on the system, it can be defined as the product of the acceleration by the total mass of the system (both cubes).
The friction force between both cubes (F) is defined as the normal force acting on the smaller cube multiplied by the coefficient of static friction. Since both cubes are subject to the same acceleration:
In order for the small cube to not slide down, the friction force must equal the weight of the small cube:
The smallest magnitude that P can have in order to keep the small cube from sliding downward is 454.11 N
Solution :
Given :
Mass attached to the spring = 4 kg
Mass dropped = 6 kg
Force constant = 100 N/m
Initial amplitude = 2 m
Therefore,
a).
= 10 m/s
Final velocity, v at equilibrium position, v = 5 m/s
Now,
A' = amplitude = 1.4142 m
b).
m' = 2m
Hence,
c).
Therefore, factor
Thus, the energy will change half times as the result of the collision.
The initial velocity is 0.65 m/s
Explanation:
The motion of the roller coaster is a uniformly accelerated motion, so we can solve the problem by using the following suvat equation:
where
s is the displacement
u is the initial velocity
v is the final velocity
t is the time
For the roller coaster in this problem:
v = 35 m/s
t = 2.3 s
s = 41 m
Solving the equation for u, we find the initial velocity:
Learn more about accelerated motion:
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