A block of inertia m is placed on an inclined plane that makes an angle theta wi

Question

A block of inertia m is placed on an inclined plane that makes an angle theta with the horizontal. The block is given a shove directly up the plane so that it has initial speed v and the coefficient of kinetic friction between the block and the plane surface is mu. How far up the plane does the block travel before it stops? Express your answer in terms of some or all of the variables m, theta, v, and mu. If the coefficient of static friction between block and surface is mu, what maximum value of theta allows the block to come to a halt somewhere on the plane and not slide back down? Express your answer in terms of some or all of the variables m, theta, v, and mu.

Explanation / Answer

A)

let a is the acceleration of the block as it slids on inclined plane.

Apply net force acting on block,

Fnet = -m*g*sin(theta) - mue*N

m*a = -m*g*sin(theta) - mue*m*g*cos(theta)

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

initial speed is v and final speed is 0.

so distance travelled before it stops,

delta_x = (vf^2 - vi^2)/(2*a)

= (0^2 - v^2)/(2*(-g*sin(theta) - mue*g*cos(theta)))

= v^2/(2*(g*sin(theta) + mue*g*cos(theta))) <<<-----Answer

B)

when static friction = compont of gravity the bock will not slide.

N*mue = m*g*sin(theta_max)

m*g*cos(theta_max))*mue = m*g*sin(theta_max)

==> sin(theta_max)/cos(theta_max) = mue

tan(theta_max) = mue

theta_max = tan^-1(mue) <<<-----Answer

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