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
The acceleration of gravity on Earth is 9.8 m/s² .
The speed of a falling object keeps increasing smoothly,
in such a way that the speed is always 9.8 m/s faster than
it was one second earlier.
If you 'drop' the penny, then it starts out with zero speed.
If you also start the clock at the same instant, then
After 1.10 sec, Speed = (1.10 x 9.8) = 10.78 meters/sec
After 1.85 sec, Speed = (1.85 x 9.8) = 18.13 meters/sec
But you want this second one given in a different unit of speed.
OK then:
= (18.13 meter/sec) x (3,600 sec/hr) x (1 mile/1609.344 meter)
= (18.13 x 3,600 / 1609.344) (mile/hr) = 40.56 mph (rounded)
We did notice that in an apparent effort to make the question
sound more erudite and sophisticated, you decided to phrase
it in terms of 'velocity'. We can answer it in those terms, if we
ASSUME that there is no wind, and the penny therefore doesn't
acquire any horizontal component of motion on its way down.
With that assumption in force, we are able to state unequivocally
and without fear of contradiction that each 'speed' described above ...
with the word 'downward' appended to it ... does become a 'velocity'.
Answer B. the direction of force is the same
