The Student Room Group

M1 Dynamics of a Particle

With workings:
1. A particle rests in limiting equilibrium on a plane inclined at 30deg to the horizontal. Determine the acceleration with which the particle will slide down the plane when the angel of inclination is increased to 40deg.

2. A particle of mass 2kg rests in limiting equilibrium on a plane inclined at 25deg to the horizontal. The angle of inclination is decreased to 20deg and a force of magnitude 20N is applied up a line of greatest slop. Find the particle's acceleration. When the particle has been moving for 2 seconds the force is removed. Determine the distance the particle will move up the plane.

3. A block of mass 1.6kg is placed on a rough plane inclined at 45deg to the horizontal. The coefficient of friction between the block and the plane is 1/4. Model the block as a particle and hence find the acceleration of the block down the plane. Find the velocity of the block after 2 seconds, assuming that it starts from rest.

Thank you! :smile:
Rep will be given :biggrin:
Trish xx
Reply 1
1) mgsin30 = F = (mu)R = (mu)mgcos30
so mu = tan30 = 1/rt3

now...

Force = ma = mgsin40 - (mu)mgcos40
a = gsin40 - (g/rt3)cos40 = 1.97ms^-2

2. A particle of mass 2kg rests in limiting equilibrium on a plane inclined at 25deg to the horizontal. The angle of inclination is decreased to 20deg and a force of magnitude 20N is applied up a line of greatest slop. Find the particle's acceleration. When the particle has been moving for 2 seconds the force is removed. Determine the distance the particle will move up the plane.

2) 2gsin25 = (mu)2gcos25
mu = tan25 = 0.466...

now friction acts in other direction...

20 - 2gsin20 - (mu)2gcos20 = Force = 2a
a = 10 - g(sin20 + tan25cos20) = 2.35ms^-2

in that time period it goes: s = ut + 1/2at^2 = 0.5 * 2.35 * 2^2 = 4.708m
velocity at end of that u + at = 4.708s.
then when it's slowing down, acceleration is g(sin20 + tan25cos20) = 7.65ms^-2, s = (v^2 - u^2)/2a = 1.45m

add them up to get 6.16m

phew. you just know there's a mistake somewhere...
Reply 2
"3. A block of mass 1.6kg is placed on a rough plane inclined at 45deg to the horizontal. The coefficient of friction between the block and the plane is 1/4. Model the block as a particle and hence find the acceleration of the block down the plane. Find the velocity of the block after 2 seconds, assuming that it starts from rest."

Reaction force = 1.6g.sin45 = 1.6g/root2
Friction = 1/4 * Reaction
= 1.6g/4root2

Resultant Force = ma
mgsin45 - 1.6g/4root2 = 1.6a
3*1.6g/4root2 = 1.6a
a = 5.2 m/s2

Velocity after 2 seconds = 5.2*2
= 10.4 m/s
Reply 3
Thank you both :smile: rep will be on the way soon :smile: