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# What is its resulting acceleration?

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?

### 3 Answers by Expert Tutors

Robert J. | Certified High School AP Calculus and Physics TeacherCertified High School AP Calculus and Ph...
4.6 4.6 (13 lesson ratings) (13)
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
Andre W. | Friendly tutor for ALL math and physics coursesFriendly tutor for ALL math and physics ...
5.0 5.0 (3 lesson ratings) (3)
1
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.