(0%) Problem 6: A block of mass m = 210 kg rests against a spring with a spring

Question

(0%) Problem 6: A block of mass m = 210 kg rests against a spring with a spring constant of k = 530 N/m on an inclined plane which makes an angle of degrees with the horizontal. Assume the spring has been compressed a distance d from its neutral position. Refer to the figure ©theexpertta.com 25% Part (a) Set your coordinates to have the x-axis along the surface of the plane, with up the plane as positive, and the y-axis normal to the plane, with out of the plane as positive. Enter an expression for the normal force, FN, that the plane exerts on the block (in the y-direction) in terms of defined quantities and g Grade Summary Potential Submissions 0% 100% cos() sin() | cos( ) sin() cos(0) sin(0) 78 9 Attempts remaining: 10 % per attempt) detailed view 0 END BACKSPACE DEL CLEAR Submit Hint I give up! Hints: 2% deduction per hint. Hints remaining: 3 Feedback: 1% deduction per feedback. 25% Part (b) Denoting the coefficient of static friction by write an expression for the sum of the forces in the x-direction Just before the block begins to slide up the inclined plane. Use defined quantities and g in your expression - 25% Part (c) Assuming the plane is frictionless, what is the minimum angle in degrees the inclined plan can make before the block will move if the spring is compressed by 0.1 m? 25% Part (d) Assuming = 45 degrees and the surface is frictionless, how far will the spring be compressed, d in meters?

Explanation / Answer

given

mass m = 210 kg , force constant k = 530 N/m

angle theta  

Part a) the normal force that the plane exerts on the block is  

F_N = mg cos theta

F_N = 210*9.8 cos theta

PArt b)

along the x axis the force exerted by the spring and the gravitational force by the mass acts  

kx cos theta - mue_s*m*g sin theta- mg sin theta = 0

Part C)

kx*cos theta - mg sin theta = 0

kx* cos theta = mg sin theta

tan theta = kx /mg

theta =arc tan (530*0.1/(210*9.8))

theta = 1.475 degrees

Partd)

by conservation of energy  

initial elastic potential energy Ui_s = 0.5*k*x^2

gravitaitonal potential energy is Ui_g = 0 J

kinetic energy of the block is kf = 0.5*m*vf^2 = 0 J

final

elastic potential energy Uf_s = 0

gravitaitonal potential energy is Uf_g = mgd sin theta J

kinetic energy of the block is ki = 0.5*m*vi^2 = 0 J

equating the initial and final total energies

o.5*k*x^2 = m*g*d sin theta

0.5*530*0.1^2 = 210*9.8*d sin(45)

solving for d

d = 0.001821 m

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