Two boxes of fruit on a frictionless horizontal surface are connected by a light string as in the figures below, where m1 = 15 kg and m2 = 17 kg. A force of 58 N is applied to the 17 kg box.
Repeat the problem for the case where the coefficient of kinetic friction between each box and the surface is 0.09.
a) find the acceleration
b) find the tension in the string
1.
The acceleration of the system is
a=F/(m1+m2) = 58/(15+17)=1.81 m/s².
T1+T2
m1•a=T
m2•a= F-T,
T = m1•a = 15•1.81=27.5 N.
2. μ = 0.09,
T1=T2=T,
m1•a=T-F1(fr)=T- μ•N1=T- μ•m1•g
m2•a=F-T-F2(fr)= F-T- μ•N2= F-T- μ•m2•g
a•(m1+m2) = T- μ•m1•g+ F-T- μ•m2•g =F- μ•g(m1+m2),
a= {F- μ•g(m1+m2)}/ (m1+m2)=
={58 – 0.09•9.8(15+17)}/(15+17) =
=0.93 m/s²
T= m1•a+ μ•m1•g=m1(a+ μ•g) =
=15(0.93+0.09•9.8)=27.18 N
To find the acceleration of the system, we can use Newton's second law, which states that the net force acting on an object is equal to the mass of the object multiplied by its acceleration.
In this case, there are two forces acting on the system: the applied force (58 N) and the force of friction. The force of friction depends on the coefficient of kinetic friction and the normal force, which is equal to the weight of each box. The weight of an object is calculated by multiplying its mass by the acceleration due to gravity (9.8 m/s^2).
a) Without friction:
Since there is no friction, the only force acting on the system is the applied force. The net force is given by:
Net Force = Applied Force - Force of Friction
Force of Friction = 0 (no friction)
Net Force = Applied Force
Net Force = 58 N
Now we can use Newton's second law to find the acceleration:
Net Force = Mass × Acceleration
58 N = (15 kg + 17 kg) × Acceleration
58 N = 32 kg × Acceleration
Dividing both sides by 32 kg, we get:
Acceleration = 58 N / 32 kg
Acceleration ≈ 1.81 m/s^2
Therefore, the acceleration of the system without friction is approximately 1.81 m/s^2.
b) With friction:
Now let's consider the case where the coefficient of kinetic friction is 0.09. We need to calculate the force of friction first using the formula:
Force of Friction = Coefficient of Kinetic Friction × Normal Force
The normal force is equal to the weight of each box. The weight of an object is given by:
Weight = Mass × Acceleration due to gravity
For the 15 kg box:
Weight1 = 15 kg × 9.8 m/s^2
Weight1 ≈ 147 N
For the 17 kg box:
Weight2 = 17 kg × 9.8 m/s^2
Weight2 ≈ 166.6 N
Now we can calculate the force of friction for each box:
Force of Friction1 = 0.09 × Weight1
Force of Friction1 = 0.09 × 147 N
Force of Friction1 ≈ 13.23 N
Force of Friction2 = 0.09 × Weight2
Force of Friction2 = 0.09 × 166.6 N
Force of Friction2 ≈ 14.94 N
The net force acting on the system is given by:
Net Force = Applied Force - Force of Friction
Net Force = 58 N - (Force of Friction1 + Force of Friction2)
Net Force = 58 N - (13.23 N + 14.94 N)
Net Force ≈ 29.83 N
Again, we can use Newton's second law to find the acceleration:
Net Force = Mass × Acceleration
29.83 N = (15 kg + 17 kg) × Acceleration
29.83 N = 32 kg × Acceleration
Dividing both sides by 32 kg, we get:
Acceleration = 29.83 N / 32 kg
Acceleration ≈ 0.93 m/s^2
Therefore, the acceleration of the system with a coefficient of kinetic friction of 0.09 is approximately 0.93 m/s^2.
Finally, to find the tension in the string, we can use the equation:
Tension = Mass × Acceleration
For the 17 kg box:
Tension = 17 kg × Acceleration
Tension ≈ 17 kg × 0.93 m/s^2
Tension ≈ 15.81 N
Therefore, the tension in the string is approximately 15.81 N.