Question

Two boxes of fruit on a frictionless horizontal surface are connected by a light string as in the figure below, where m1 = 16 kg and m2 = 27 kg. A force of F= 74 N is applied to the 27-kg block.

(a) Determine the acceleration of each block and the tension in the string. acceleration of m₁m/s²acceleration of m₂m/s² tension in the string N(b) Repeat the problem for the case where the coefficient of kinetic friction between each block and the surface is 0.10.

acceleration of m1	m/s2
acceleration of m2	m/s2
tension in the string	N

Answer 1

Given data:

the masses m1 = 16 kg , m2 = 27 kg
force F = 74 N
(a)
According to Newton's second law of motion,
F = (m1+m2)a

==> a = F/(m1+m2)
a = 74/43
   a = 1.720 m/s^2
acceleration of m1 is 1.720 m/s^2
acceleration of m2 is 1.720 m/s^2
the tension in the string , T = m1a

= 16*1.720

= 27.534 N

(b)

if the friction is acting on the surface, μ = 0.10
     Applying the Newton 2nd law of motion,
   Σ F = F-f1-f2
where f1 = μm1g =0.10*16*9.8 = 15.68 N
f2 = 0.10*27*9.8 = 26.46 N
therefore    Σ F = 74 - 42.14
                         = 31.86 N
the acceleration , a = ΣF/(m1+m2)

= 0.740 m/s^2
acceleration of m1 is 0.740 m/s2
acceleration of m2 is 0.740 m/s2
the tension, T = m1a + f1
                 T = 11.85 + 15.68
                 T = 27.53 N