A 2.09 kg block is placed on top of a 10.9 kg
block. A horizontal force of F = 63.3 N is
applied to the 10.9 kg block, and the 2.09 kg
block is tied to the wall. The coefficient of
kinetic friction between all moving surfaces
is 0.117. There is friction both between the
masses and between the 10.9 kg block and the
ground.
The acceleration of gravity is 9.8 m/s2 .
2.09 kg
10.9 kg
ì = 0.117
ì = 0.117
F
T
Determine the tension T in the string.
Answer in units of N.
To determine the tension T in the string, we can start by analyzing the forces acting on the system.
1. Weight force (mg):
The weight force acting on the 2.09 kg block is given by:
F1 = m1 * g = 2.09 kg * 9.8 m/s^2 = 20.422 N
The weight force acting on the 10.9 kg block is given by:
F2 = m2 * g = 10.9 kg * 9.8 m/s^2 = 106.82 N
2. Frictional forces:
The coefficient of kinetic friction between all moving surfaces is given as 0.117.
For the 2.09 kg block, the frictional force is:
F1_friction = µ * F1 = 0.117 * 20.422 N = 2.3914 N
For the 10.9 kg block, the frictional force is:
F2_friction = µ * F2 = 0.117 * 106.82 N = 12.5054 N
3. Applied horizontal force (F):
The horizontal force applied to the 10.9 kg block is given as F = 63.3 N.
Now, let's analyze the forces acting on the 10.9 kg block:
Net horizontal force on the 10.9 kg block:
F_net = F - F2_friction = 63.3 N - 12.5054 N = 50.7946 N
Since the 2.09 kg block is tied to the wall, the tension T in the string is the force required to accelerate the 2.09 kg block and overcome the frictional force on the 10.9 kg block.
To find the tension T, we can write the equation of motion for the 2.09 kg block:
T - F1_friction = m1 * a
Since the 2.09 kg block is not accelerating (a = 0), we can rewrite the equation as:
T = F1_friction = 2.3914 N
Therefore, the tension T in the string is 2.3914 N.
To determine the tension T in the string, we need to analyze the forces acting on the system.
Let's break down the forces on the 10.9 kg block first:
1. The force F = 63.3 N applied horizontally.
2. The weight (mg) acting vertically downwards, where m = mass of the 10.9 kg block and g = acceleration due to gravity (9.8 m/s²).
3. The frictional force (μN) opposing the motion, where μ = coefficient of kinetic friction and N = normal force.
Now, let's calculate each force:
1. The force F = 63.3 N applied horizontally remains unchanged.
2. The weight (mg) = (10.9 kg)(9.8 m/s²) = 106.82 N acting vertically downwards.
3. The normal force N is equal to the weight of the block on top. Since the 2.09 kg block is tied to the wall, it does not exert a vertical force on the 10.9 kg block. Thus, N = weight of the 2.09 kg block = (2.09 kg)(9.8 m/s²) = 20.39 N.
Therefore, the frictional force opposing the motion is μN = (0.117)(20.39 N) = 2.389 N.
Next, we can determine the net force acting on the 10.9 kg block:
Net force = Force applied - Frictional force
= F - μN
= 63.3 N - 2.389 N
= 60.911 N
Since the system is connected by a string, the tension T in the string will be equal to the net force acting on the 10.9 kg block:
T = Net force = 60.911 N
Therefore, the tension T in the string is 60.911 N.