What is its resulting acceleration?

Question

Two 5 kg masses are attached to opposite ends of a long massless cord which passes tautly over a massless frictionless pulley. The upper mass is initially held at rest on a table 50 cm from the pulley. The coefficient of kinetic friction between this mass and the table is 0.2. When the system is released, what is its resulting acceleration?

 

Answer: 3.9 m/s^2

Answer 1

Treat the two masses as one system. The downward force is mg, and the friction force is mgμ.

 

Apply Newton's second law to the system:

mg - mgμ = (m+m)a

Solve for a,

a = (1/2)g(1-μ) = (1/2)(9.8)(1-0.2) = 3.92 m/s^2 <==Answer

Answer 2

Apply Newton's 2nd law to each mass.

 

For the upper mass, tension and friction are acting in opposite directions and the mass moves in the direction of the tension force, so that

 

T-µmg = ma.

 

For the hanging mass, tension and gravity are acting in opposite directions and the mass moves in the direction of gravity, so that

 

T-mg = -ma.

 

Eliminate tension from the two equations by solving both for T:

 

T = ma+µmg = mg-ma.

 

Divide this by m and solve for a:

 

a+µg = g-a,

 

a=g(1-µ)/2 = 9.8(0.8)/2 = 3.92 m/s².

Answer 3

Hey Sun -- the gravity drive is 50N ... the opposing friction is 10N (20% of wt) ...

 

40N net moving 10kg total ==> accel 4m/s/s ... All the best, sir :)