mass= 2.0kg k= 40 N/m Angle (theta)= 30 degrees The block in the figure is initi
ID: 2114931 • Letter: M
Question
mass= 2.0kgk= 40 N/m
Angle (theta)= 30 degrees
The block in the figure is initially at rest on an inclined plane at the equilibrium position that it would have if there were no friction between the block and the plane. How much work is required to move the block 15 cm down the plane if the frictional coefficient is= 0.? Answer to this part is: 4.5X10^-1...how do i do the next part below?
How much work is required to move the block 15 cm down the plane if the frictional coefficient is ? = 0.17?
This may help:
For the first part the equation is W= (1/2)k(x1^2-x0^2)-m*g*sin(theta)*distance. The x values are found using x0=mgsin(theta)/K x1= xo distance The second part is just inserting mu, W= (1/2)k(x1^2-x0^2)- mu*m*g*cos(theta)*distance The reason you use cos(theta) is because you are now trying to find the value for the x axis. but i still cant get the right answer!
1 Incorrect. (Try 1) 1.487 J
2 Incorrect. (Try 2) 1.054 J
4 Incorrect. (Try 3) 1.42 J
5 Incorrect. (Try 4) 4.02 J
6 Incorrect. (Try 5) 2.13 J
7 Incorrect. (Try 6) 0.11 J
NONE OF THE ABOVE ANSWERS ARE CORRECT....
Explanation / Answer
F=mgsin30-uN = 0.71 N W=F.x= 0.71 x 15 x 10^-2 = 0.11 J
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