A 195 g block is pressed against a spring of force constant 1.60 kN/m until the block compresses the spring 10.0 cm. The spring rests at the bottom of a ramp inclined at 60.0° to the horizontal. Using energy considerations, determine how far up the incline (in m) the block moves from its initial position before it stops under the following conditions(a) if the ramp exerts no friction force on the block(b) if the coefficient of kinetic friction is 0.440 Tm(c)What If? If the ramp is 4.00 m long, what is the maximum coefficient of friction that would allow the block to reach the end of the ramp?

Energy Kinetic is the energy of the spring so 'K' is the force constant and xk is the distance the block compresses the spring and 'h' is the height for a ramp 60 angle so:

a).

[tex]\frac{1}{2}*1.60\frac{kN}{m}*(0.1m)^2=0.195kg*9.8\frac{m}{s^{2}}*d*sin(60)[/tex]

Solving for 'd'

[tex]d=\frac{1.60\frac{kN}{m}*(0.1m)^2}{2*0.195kg*9.8\frac{m}{s^2}*sin(60)}[/tex]

[tex]d=4.83m[/tex]

b).

Now the force of friction is acting so the Ep add the force

[tex]E_{k}=E_{p}[/tex]

[tex]\frac{1}{2}*K*x_{k}=m*g*h+u*m*g*h[/tex]

[tex]\frac{1}{2}*k*x_{k}^2=m*g*d*sin(60)+u*m*g*d*cos(60)[/tex]

Get the factor to resolve d

[tex]\frac{1}{2}*k*x_{k}=m*g* d*(sin(60)+u*cos(60))[/tex]

[tex]d=\frac{k*x_{k}^2}{2*m*g*(sin(60)+u*cos(60))}[/tex]

[tex]d=\frac{1.60kN/m*(0.1m)^2}{2*0.195kg*9.8 m/s^2*(1.08)}[/tex]

[tex]d=3.876m[/tex]

c).

Finally if the distance of the rampo is knowing determinate the u coefficient of friction can be find

[tex]\frac{1}{2}*k*x_{k}^2=m*g*d*sin(60)+u*m*g*d*cos(60)[/tex]

[tex]u=\frac{k*x_{k}^2}{2*m*d*cos(60)}-tan(60)[/tex]

[tex]u=\frac{1.6kN/m*(0.1)^2}{2*0.195kg*4m*cos(60)}-tan(60)[/tex]

[tex]u=0.361[/tex]

annyksl annyksl · Answer 2 · 2022-04-07T13:36:06+02:00

The distance the block will travel without the friction force will be 4.83 meters. The distance traveled with the coefficient of friction will be equal to 3.876 meters and if the ramp is 400 meters long, the maximum coefficient of friction will be equal to 0.36.

How can we arrive at this result?

To find the distance traveled by the block without friction force, we will use the equation:

[tex]0.5*K*X_k=mgh\\[/tex]

Using the values presented in the question, we can substitute them into the equation as follows:

[tex]0.5*1.60*(0.1)^2=0.195*9.8*d*sin(60)\\d=\frac{0.5*1.60*(0.1)^2}{0.195*9.8*sin(60)} \\d= 4.83 m[/tex]

To calculate the distance traveled using the indicated friction force, we will use the equation:

[tex]d= \frac{K*x^2_k}{2*m*g*sin(60)+u*cos(60)}[/tex]

Using the values presented in the question, we can solve the equation as follows:

[tex]d=\frac{1.60*(01)^2}{2*0.195*9.8*(1.08)} = 3.876m[/tex]

Finally, we can calculate the maximum coefficient of friction on a 4.00 meter ramp with the following equation:

[tex]u=\frac{K*x^2_k}{2*m*d*cos (60)} - tan (60)\\u= \frac{1.6*(0.1)^2}{2*0.195*4*cos(60)} -tang(60)\\ = 0.361[/tex]

More information about friction force is in the link:

https://brainly.com/question/13680415