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
a) Well, let's see. If there is no friction between the boxes and the surface, then the only force acting on the system is the applied force of 58 N. Since the mass of the system is the sum of m1 and m2 (15 kg + 17 kg), we can calculate the acceleration using Newton's second law (F = ma). So, a = F/m = 58 N / (15 kg + 17 kg) = 1.44 m/s².
b) Now, let's consider the case where there is some friction present. With a coefficient of kinetic friction of 0.09, we can calculate the frictional force acting on each box by multiplying the coefficient by the normal force (which is just the weight of the box, since they are on a horizontal surface). The normal force can be calculated by multiplying the mass of each box by the acceleration due to gravity (9.8 m/s²). So, for both boxes, the frictional force is 0.09 * (mass * 9.8 m/s²).
For the 15 kg box, the frictional force is 0.09 * (15 kg * 9.8 m/s²) = 13.23 N.
And for the 17 kg box, the frictional force is 0.09 * (17 kg * 9.8 m/s²) = 15.83 N.
Now, let's take into account the applied force of 58 N. In the horizontal direction, the net force is the difference between the applied force and the sum of the frictional forces. So, the net force is 58 N - (13.23 N + 15.83 N) = 29.94 N.
Using Newton's second law again, we can calculate the acceleration: a = F/m = 29.94 N / (15 kg + 17 kg) = 0.75 m/s².
So, in this case, the acceleration is 0.75 m/s² and the tension in the string can be found using the mass of the 17 kg box and the acceleration: T = m2 * a = 17 kg * 0.75 m/s² = 12.75 N.
Hope that helps! And remember, friction can really slow things down, just like trying to walk in clown shoes on a banana peel!
To solve this problem, we'll use Newton's second law of motion, which states that the total force acting on an object is equal to the mass of the object multiplied by its acceleration.
a) Without friction:
For the case without friction, the only force acting on the system is the force applied to the 17 kg box, which is 58 N. Therefore, the total force on the system is the same as the force applied to the 17 kg box.
The equation can be written as:
F = m1a + m2a
where F is the force applied to the 17 kg box, m1 is the mass of the 15 kg box, m2 is the mass of the 17 kg box, and a is the acceleration of the system.
Substituting the given values:
58 N = (15 kg)a + (17 kg)a
58 N = 32a
a = 58 N / 32 kg
a ≈ 1.81 m/s^2
So the acceleration of the system is approximately 1.81 m/s^2.
b) With friction:
In this case, we need to consider the frictional force acting on each box. The equation for the frictional force can be written as:
f_friction = μ * N
where μ is the coefficient of kinetic friction and N is the normal force.
The normal force on each box can be calculated as:
N = m * g
where m is the mass of the box and g is the acceleration due to gravity (approximately 9.8 m/s^2).
For the 15 kg box:
N1 = (15 kg) * (9.8 m/s^2)
N1 ≈ 147 N
For the 17 kg box:
N2 = (17 kg) * (9.8 m/s^2)
N2 ≈ 166.6 N
Now we can calculate the frictional force on each box:
f_friction1 = (0.09) * (147 N)
f_friction1 ≈ 13.23 N
f_friction2 = (0.09) * (166.6 N)
f_friction2 ≈ 14.99 N
The total force acting on the 15 kg box is:
F1 = -f_friction1
F1 ≈ -13.23 N
The net force acting on the 17 kg box is:
F2 = 58 N - f_friction2
F2 ≈ 58 N - 14.99 N
F2 ≈ 43.01 N
Now we can write down the equation of motion for the system:
F2 - F1 = (m1 + m2) * a
Substituting the values:
43.01 N - (-13.23 N) = (15 kg + 17 kg) * a
56.24 N = 32 kg * a
a = 56.24 N / 32 kg
a ≈ 1.76 m/s^2
So the acceleration of the system, considering the friction, is approximately 1.76 m/s^2.
To find the tension in the string, we can use the equation for the net force on the 17 kg box:
F2 = m2 * a + f_tension
where f_tension is the tension in the string.
Substituting the values:
43.01 N = (17 kg) * (1.76 m/s^2) + f_tension
43.01 N = 29.92 N + f_tension
f_tension = 43.01 N - 29.92 N
f_tension ≈ 13.09 N
So the tension in the string is approximately 13.09 N.
To solve this problem, we will use Newton's second law of motion, which states that the net force acting on an object is equal to the product of its mass and acceleration:
F_net = m * a
where F_net is the net force, m is the mass of the object, and a is the acceleration.
a) Finding the acceleration:
In the first case, where there is no friction between the boxes and the surface, the only horizontal force acting on the system is the applied force of 58 N. Since the two boxes are connected by a light string, they will have the same acceleration. We can find the acceleration by using the equation mentioned above:
F_net = (m1 + m2) * a
58 N = (15 kg + 17 kg) * a
58 N = 32 kg * a
a = 58 N / 32 kg
a = 1.8125 m/s^2
Therefore, the acceleration of the system in the first case is 1.8125 m/s^2.
b) Finding the tension in the string:
In the case where there is no friction, the tension in the string is the same as the force applied to the 17 kg box. So, the tension T is 58 N.
Now let's consider the second case, where there is a coefficient of kinetic friction between each box and the surface. In this case, we need to account for the frictional force.
The frictional force can be calculated using the equation:
F_friction = μ * m * g
where μ is the coefficient of kinetic friction, m is the mass of the box, and g is the acceleration due to gravity (approximately 9.8 m/s^2).
For the 15 kg box:
F_friction1 = 0.09 * 15 kg * 9.8 m/s^2
F_friction1 = 13.23 N
For the 17 kg box:
F_friction2 = 0.09 * 17 kg * 9.8 m/s^2
F_friction2 = 15.771 N
Now, let's find the net force:
F_net2 = (m1 + m2) * a
F_net2 = (15 kg + 17 kg) * a
F_net2 = 32 kg * a
Since there is a force of 58 N applied to the 17 kg box, the net force can be calculated as:
F_net2 = 58 N - F_friction2
F_net2 = 58 N - 15.771 N
F_net2 = 42.229 N
Now we can find the acceleration:
42.229 N = 32 kg * a
a = 42.229 N / 32 kg
a = 1.3190 m/s^2
Therefore, the acceleration of the system in the second case is 1.3190 m/s^2.
To find the tension in the string, we can use the same approach as in the first case. Since the boxes are connected by a light string, the tension in the string for the second case is equal to the net force acting on the 17 kg box:
T = F_net2
T = 42.229 N
Therefore, the tension in the string for the second case is 42.229 N.