physics help needed question:object #1 slides along a surface (μk = 0.15) that makes a 20º angle horizontal.

Question

object #1 slides along a surface (μk = 0.15) that makes a 20º angle with respect to the horizontal. Both objects are connected by a massless string that runs over a frictionless pulley (m = 10 kg, r = 0.25 m). Object #1 has a mass of 5 kg and object #2 has a mass of 20 kg. Apply Newton’s Second Law to each of the three objects to obtain a system of three equations and three unknows that can be solved to find the tensions and acceleration of this system. Then calculate the final speed of object #2 after dropping a distance of 2 m.

Answer 1

This is a classic illustration of the treatment of inertia in the pulley system and accounting for acceleration in the weight of an object.

Make a free body diagram of the pulley

it has T2, which is the tension in the string from the pulley to object 2 which has mass m2

T2=m2*g-m2*a
where a is the acceleration of the mass downward

It has T1, which is the tension in the string from the pulley to object 1 which has mass m1
T1=sin(20)*m1*g
+μk*cos(20)*m1*g
+m1*a

Now sum the torques about the center of the pulley:

R*T2-R*T1=I*q
where q is the angular acceleration

for a solid disk,
I=1/2*mp*R^2
where mp is the mass of the pulley

q=a/R

There's your three equations:

T2=m2*g-m2*a

T1=sin(20)*m1*g
+μk*cos(20)*m1*g
+m1*a

R*T2-R*T1=
1/2*mp*R*a

Note that R divides out
T2-T1=1/2*mp*a

I computed
a=5.75 m/s^2

Since you can solve for a using the above equations,
then the speed of object#2 after dropping 2m is:

Sqrt(a*d)/2
where d=2m

v=1.7 m/s

j

Answer 2

F = ma down plane gives mgsinA - F = ma where F = (mu)N N = mgcosA So mgsinA - (mu)mgcosA = ma and if a = 0 then mgsinA - (mu)mgcosA = 0 sinA - (mu)cosA = 0 tanA = (mu) which is a well-known result. If mu = 1 then A = 45 degrees Hope this helps.

Answer 3

Where is the pulley? At the upper or at the lower end of the slope? Is the brick sliding up or down? It’s important kid!