Question

B. Block on an incline

Answer 1

Block on the incline:

• net force parallel to the incline

∑ F (para.) = m₁ g sin(38°) - T = m₁ a

where T is the magnitude of tension in the cord.

Notice that we take down-the-incline to be the positive direction, so that if the 3.9-kg block pulls the 2.6-kg block upwards, then the acceleration of the system is positive.

Suspended block:

• net vertical force

∑ F (vert.) = T - m₂ g = m₂ a

Solve both equations for the acceleration a, set the results equal to one another, and solve for T :

a = g sin(38°) - T/m₁

a = T/m₂ - g

==>   g sin(38°) - T/m₁ = T/m₂ - g

==>   T (1/m₂ + 1/m₁) = g (sin(38°) + 1)

==>   T = g (sin(38°) + 1) / (1/m₂ + 1/m₁)

==>   T = (9.81 m/s²) (sin(38°) + 1) / (1/(2.6 kg) + 1/(3.9 kg)) ≈ 25 N