Chapter 5 H : 6/15/2016 11:3000 PM (17%) Problem 5: A block of mass m 120 kg con

Question

Chapter 5 H : 6/15/2016 11:3000 PM (17%) Problem 5: A block of mass m 120 kg constant of k 750 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 m=120 kg k = 750 N/m Otheexpertfa.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. Write an expression for the normal force. Fs. in the y-direction. Ponestial 100% Subnissions l per atempt) sin() | sin(O) 0 5

Explanation / Answer

a) in the y direction the normal force will be given by M*g*cos*theta

b) as the spring is already compressed some distance d the block will have a force acting in the direction of action of friction.at no motion these will be balanced by the gravity force acting in x direction.

hence we will have,

M*g*sin(theta) = k*d + mu*M*g

where K is he spring constant

g is the acceleation due to gravity.

theta is the agle of inclination

M is the Mass of the of the block

c) we use the aove formula ,where mu =0

so,

M*g*sin(theta) = k*d

=>120*9.8 * sin(theta) = 750*0.1

=> theta = 3.656 degrees

d) again we use the same formula but now the unknown value is d

so we have,

M*g*sin(theta) = k*d

=>120*9.8*sin(45) = 750*d

=> d= 1.1087 m

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