Mass on Inclined Plane This exercise uses a simulation which

Mass on Inclined Plane. This exercise uses a simulation which is not included. In the simulation, a massive block is situated on an inclined plane. But, Please help answer the following questions given the information. If more information is needed from the simulation, please let me know.

A. mass = 10 kg and angle = 30 degrees. Plot Vy versus Vx. (Remember that means y axis against x axis). Set both axes to 10 m/s.
Why is the plot a straight line? ____________ Give an expression for the slope of this line in terms of trigonometric functions:______________.

B. angle = 27 degrees and coefficient of friction = 0.5. Plot Vx versus time. What is the value for ax? ________ m/s^2

C. angle = 20 degrees and coefficient of fricion = 0.20. Set ymax = 500 J. Run the simulation. After total E, KE, or PE (depending on graph plotted) reaches a stable value, stop the simulation and print the graph. Answer the following questions accordingly:

I. Plot total E versus time, t. What do you think happens to the energy that is lost from the block? Notice the shape of the curve.

II. Plot kinetic energy KE vs. time, t. Compare the graph to the graph of total energy. Explain what happens to E and KE just before and just after the block \"bounces.\"

III. Plot potential energy PE versus time, t. Compare this graph to the graph of total energy. Explain what happens to E and PE just before and just after the block \"bounces\".

Solution

Ans:

m= 10 kg angle = 30⁰

Force on the object = mg

resolve the forces along the plane and ^_r

F_x = mgSin()

F_{y =} mgCos()

acceleration a = F/m

a_x = g Sin()

a_y = g Cos()

v= at

v_x = a_xt, v_y=a_yt

v_y = v_x(a_x/a_y) = tan() v_x

slope of the curve tan() is constant for a any given inclined plane, hence the curve is always a straight line.

Normal force F_y = mgCos()

frictional force f_s = F_y

Force along the plane F_x = mgSin()

net force down the plane F_n = mg(Sin()- Cos())

acceleration a_x = g(Sin()- Cos())

= 9.8(Sin(27) -0.5*Cos(27)) = 0.0085 m/s²

velocity along the plane

           v_x = a_x t = (Sin()- Cos())gt

                   = (Sin(27) -0.5*Cos(27)) t

                   = 0.083t

= 20⁰ , = 0.2

f_s = 0.2*10*9.8*Cos(20) = 18.42 N

v_x = (Sin(20)- 0.2 Cos(20))9.8t = 1.51t

v_y = 9.8Cos(20)t = 9.21t

KE = 0.5mv² = 0.5*10*(1.51t)² + (9.21t)² = 435.52t²

   = change in PE - work done for friction

work done for friction f_s *v_xt = 18.42*1.51t = 27.82t

total energy E = 435.52t² + 27.82 t

PE = mg(0.5a_yt²) = 10*9.8*(0.5* 9.8* Cos(20)t²) = 451.24t²

loss of E, E is used to work against the frictional force, it is work done against frictional force F_y