please help! SAME (111 A block of mass M = 10 kg is in motion downhill on a fric

Question

please help!

SAME (111 A block of mass M = 10 kg is in motion downhill on a frictionless plane inclined at 20° from the horizontal AM (a) Compute the coefficient of static friction which would correspond to an angle of repose equal to 20 (b) Determine the accleration of the block down the incline. (c) Determine the normal force of contact acting between the block and the incline. (d) (G) Suppose that kinetic friction acts between the incline and the block, with Ik Compute the acceleration of the block down the incline in this case. 0.2. (i) Suppose that the coefficient of kinetic friction is0.4, and compute the accelera- tion of the block in this case (e) Instead suppose that the mass of the block is 20kg. Choose the factor by which the results for (b) (d) i-) change. (b) acceleration 1/ 1/2 1 2 4 other (c) normal force1/4 1/2 2 4other (d) (i) acceleration1/4 1/2 2 other (d) (ii) acceleration 1/4 1/2 2 4 other

Explanation / Answer

According to cheggs policy, I will answer only the first 4 parts.

a)The force on the body along the incline = M*g*sin(20)

let the coefficient of static friction be u.

so, uMg = Mgsin(20)

or u=sin(20)

=0.34

b)acceleration on the block down the incline = force/Mass

=Mg*sin(20)/M

=g*sin(20)

=9.81*sin(20)

=3.355 m/s^2

c)Normal force = Mgcos(20)

=10*9.81*0.939

=92.1159 N

d)let the acceleration be a.so,

Gravitational force - fricitonal force = M*a

or Mg*sin(20) - u*N = Ma

or 3.355 - 0.2*9.81*0.939 = a (since N = M*g*cos(20))

or a=1.51 m/s^2

Bonus answers, Please rate the answer

if uk=0.4,

3.355 - 0.4*9.81*0.939 = a

or a=-0.329

since a is negative, the block will not move.

e)Since only force is dependent on Mass,

only the normal force will change by a factor of 2

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