The acceleration of a particle along a straight line is defined by a=(2t−9m/s2 where t is in seconds. when t=9 determine the par
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Answer:
5.7141 m
Explanation:
Here the potential and kinetic energy will balance each other
This is the initial velocity of the system and the final velocity is 0
t = Time taken = 0.04 seconds
F = Force = 18000 N
a = Acceleration
g = Acceleration due to gravity = 9.81 m/s²
Equation of motion
From Newton's second law
Squarring both sides
The height from which the student fell is 5.7141 m
Answer:
6.57 m/s
Explanation:
First use Hook's Law to determine the F the compressed spring acts on the mass. Hook's Law F=kx; F=force, k=stiffnes of spring (or spring constant), x=displacement
F=kx; F=180(.3) = 54 N
Next from Newton's second law find the acceleration of the mass.
Newton's .2nd law F=ma; a=F/m ; a=54/.75 = 72m/s²
Now use the kinematic equation for velocity (or speed)
v₂²= v₀² + 2a(x₂-x₀); v₂=final velocity; v₀=initial velocity; a=acceleration; x₂=final displacement; x₀=initial displacment.
v₀=0, since the mass is at rest before we release it
a=72 m/s² (from above)
x₀=0 as the start position already compressed
x₂=0.3m (this puts the spring back to it's natural length)
v₂²= 0 + 2(72)(0.3) = 43.2 m²/s²
v₂= = 6.57 m/s