Answer:
0.1 N
Explanation:
Considering the relationship between force,
spring constant and extension as defined by Hook's law
The force F=xk as from Hooke's law where F is the force of the spring, k is spring constant and x is extension or compression. Substituting 2 N/m for k and 5cm which is equivalent to 0.05 m for extention x then the force will be
F=2*0.05=0.1 N
Answer:
a) T=1.35s
b) amplitude = 0.0923m
Explanation:
m=300 gr
k=6.5 N/m
first we need to get the angular frequency of the motion
so we have that
ω = √(k/m)
in this case motion is a simple harmonic so the period is defined by:
T= 2π / ω
T= 2π / √(k/m)
replacing the variables...
T= 2π / √(6.5/0.3)
T=1.35s (period of the block's motion)
and...
α max = | ω²r max |
2 = (2π/1.35)² * r max
r max= 0.0923m
We need to see what forces act on the box:
In the x direction:
Fh-Ff-Gsinα=ma, where Fh is the horizontal force that is pulling the box up the incline, Ff is the force of friction, Gsinα is the horizontal component of the gravitational force, m is mass of the box and a is the acceleration of the box.
In the y direction:
N-Gcosα = 0, where N is the force of the ramp and Gcosα is the vertical component of the gravitational force.
From N-Gcosα=0 we get:
N=Gcosα, we will need this for the force of friction.
Now to solve for Fh:
Fh=ma + Ff + Gsinα,
Ff=μN=μGcosα, this is the friction force where μ is the coefficient of friction. We put that into the equation for Fh.
G=mg, where m is the mass of the box and g=9.81 m/s²
Fh=ma + μmgcosα+mgsinα
Now we plug in the numbers and get:
Fh=6*3.6 + 0.3*6*9.81*0.8 + 6*9.81*0.6 = 21.6 + 14.1 + 35.3 = 71 N
The horizontal force for pulling the body up the ramp needs to be Fh=71 N.
