what is the answer about e and f ? thank you 12. 1.44/3.42 paints | Previcus My

Question

what is the answer about e and f ? thank you

12. 1.44/3.42 paints | Previcus My Notes Ask Your Teacher A 10.0 kg block on an incline plane of vanable angle. The coefficients of static and kinetic friction are ,-0.25 and ,-0. i 8. (a) What is the minimum angle such that the block slides down the plane? 14 (b) What is the frictional force at this angle? (c) If the angle of the incline plane is half of that in part (a) what are the force of friction and acceleration of the block down the plane if it relcased from rest? 1 9 m/s2 (d) If the angle of the incline plane is twice that in part (a) what are the torce ot friction and acceleration ot the block down the plane if it released from rest? 15.6 304s

Explanation / Answer

e) theta = 14/2 = 7 degrees

fk = mue_k*N

= mue_k*m*g*cos(theta)

= 0.18*10*9.8*cos(7)

= 17.5 N <<<<<<<--------Answer

a = -g*sin(theta) - mue_k*g*cos(theta)

= -9.8*sin(7) - 0.18*9.8*cos(7)

= -2.94 m/s^2

|a| = 2.94 m/s^2 <<<<<<<--------Answer

after reaching maximum height,
fs = mue_s*N

= mue_s*m*g*cos(theta)

= 0.25*10*9.8*cos(7)

= 24.3 N <<<<<<<--------Answer

a = 0 <<<<<<<--------Answer

since static friction is grater than component of gravity.

f)

theta = 14*2 = 28 degrees

fk = mue_k*N

= mue_k*m*g*cos(theta)

= 0.18*10*9.8*cos(28)

= 15.6 N <<<<<<<--------Answer

a = -g*sin(theta) - mue_k*g*cos(theta)

= -9.8*sin(28) - 0.18*9.8*cos(28)

= -6.16 m/s^2

|a| = 6.16 m/s^2 <<<<<<<--------Answer

after reaching maximum height,
fs = mue_s*N

= mue_s*m*g*cos(theta)

= 0.25*10*9.8*cos(14)

= 23.8 N <<<<<<<--------Answer

a = g*sin(theta) - mue_s*g*cos(theta)

= 9.8*sin(28) - 0.25*9.8*cos(28)

= 2.44 m/s^2 <<<<<<<--------Answer

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