A bumper cart has a mass of 200 kg and has a protective bumper around it that behaves like a spring. The spring constant is 5000
1 answer:
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Answer:
23.52092 J
Explanation:
m = Mass of block = 6.79 kg
s = Sliding distance = 2.82 m
= Angle of slide = 20.7°
= Coefficient of kinetic friction = 0.425
g = Acceleration due to gravity = 9.8 m/s²
Work done by the force of gravity is given by
The work done by the force of gravity is 23.52092 J
Answer:
Approximately , assuming friction between the vehicle and the ground is negligible.
Explanation:
Let denote the mass of the vehicle. Let
denote the initial velocity of the vehicle. Let
denote the spring constant (needs to be found.) Let
denote the maximum displacement of the spring.
Convert velocity of the vehicle to standard units (meters per second):
.
Initial kinetic energy () of the vehicle:
.
When the vehicle is brought to a rest, the elastic potential energy () stored in the spring would be:
.
By the conservation of energy, if the friction between the vehicle and the ground is negligible, the initial of the vehicle should be equal to the
of the vehicle. In other words:
.
Rearrange this equation to find an expression for , the spring constant:
.
Substitute in the given values ,
, and
: