Question

A 2.00-kg object A is connected with a massless string across a massless, frictionless pulley to a 3.00-kg object B. Object A rests on a nearly frictionless plane, which is tilted at an angle of 40.0∘ as shown.

Answer 1

Answer:

• tension: 19.3 N
• acceleration: 3.36 m/s^2

Explanation:

Given

  mass A = 2.0 kg

  mass B = 3.0 kg

  θ = 40°

Find

  The tension in the string

  The acceleration of the masses

Solution

Mass A is being pulled down the inclined plane by a force due to gravity of ...

  F = mg·sin(θ) = (2 kg)(9.8 m/s^2)(0.642788) = 12.5986 N

Mass B is being pulled downward by gravity with a force of ...

  F = mg = (3 kg)(9.8 m/s^2) = 29.4 N

The tension in the string, T, is such that the net force on each mass results in the same acceleration:

  F/m = a = F/m

  (T -12.59806 N)/(2 kg) = (29.4 N -T) N/(3 kg)

  T = (2(29.4) +3(12.5986))/5 = 19.3192 N

__

Then the acceleration of B is ...

  a = F/m = (29.4 -19.3192) N/(3 kg) = 3.36027 m/s^2

The string tension is about 19.3 N; the acceleration of the masses is about 3.36 m/s^2.