What is the potential energy of the two-spring system after the point of connection has been moved to position (x, y)? keep in m
2 answers:
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Answer:
Velocity component along y-axis is 2 m/s.
Explanation:
Given that
Differentiating (1) w.r.to t
Velocity component along y-axis is 2 m/s.
Answer:
Magnitude of the force is 2601.9 N
Explanation:
m = 450 kg
coefficient of static friction μs = 0.73
coefficient of kinetic friction is μk = 0.59
The force required to start crate moving is .
but once crate starts moving the force of friction is reduced .
Hence to keep crate moving at constant velocity we have to reduce the force pushing crate ie .
Then the above pushing force will equal the frictional force due to kinetic friction and constant velocity is possible as forces are balanced.
Magnitude of the force
To solve this problem we will use the Force equation according to the definition given in Newton's second law. There we have that the Force is equal to
Where,
m = mass
a = acceleration
Our values are given as
Considering that both mass are equal to one, we have that:
Therefore the truck exert a force on the car of 11kN