To solve this problem, we need to analyze the forces acting on the block as it moves between the two vertical walls. Given that the block has a mass of 1 kg and the coefficient of kinetic friction (μk) is 0.5, we can start by determining the forces involved and how they interact with the motion of the block.
Understanding the Forces at Play
When the block moves between the walls, it experiences several forces:
- Weight (W): This is the force due to gravity acting downwards, calculated as W = mg, where m is the mass and g is the acceleration due to gravity (approximately 9.81 m/s²).
- Normal Forces (N): Each wall exerts a normal force on the block. If we denote the normal force from the left wall as N1 and from the right wall as N2, these forces act perpendicular to the walls.
- Frictional Forces (F_f): The frictional force opposes the motion of the block and is given by F_f = μk * N, where N is the normal force exerted by the walls.
Calculating the Weight of the Block
First, let's calculate the weight of the block:
W = mg = 1 kg * 9.81 m/s² = 9.81 N
Determining the Motion of the Block
Since the block starts from rest and travels a distance of 1 meter in 1 second, we can find its acceleration using the kinematic equation:
d = ut + (1/2)at²
Here, d = 1 m, u = 0 (initial velocity), and t = 1 s. Plugging in the values:
1 = 0 + (1/2)a(1)²
1 = (1/2)a
Thus, a = 2 m/s².
Applying Newton's Second Law
According to Newton's second law, the net force acting on the block in the horizontal direction (x-direction) can be expressed as:
F_net = ma
Since the block is moving horizontally between the walls, the net force is also the difference between the frictional forces exerted by the two walls:
F_net = F_f1 - F_f2
Assuming the normal forces from both walls are equal (N1 = N2 = N), we can express the frictional forces as:
F_f1 = μk * N and F_f2 = μk * N
Thus, we have:
F_net = μk * N - μk * N = 0
This indicates that the frictional forces balance each other out, and we need to consider the vertical forces acting on the block.
Vertical Forces and Normal Forces
The vertical forces acting on the block include its weight and the normal forces from the walls. Since the block is not accelerating vertically, the sum of the vertical forces must equal zero:
N1 + N2 - W = 0
Substituting W:
N1 + N2 = 9.81 N
Finding the Contact Force Exerted by Each Wall
Assuming the normal forces from both walls are equal (N1 = N2 = N), we can rewrite the equation as:
2N = 9.81 N
Thus, N = 9.81 N / 2 = 4.905 N.
Calculating the Frictional Force
Now, we can find the frictional force exerted by each wall:
F_f = μk * N = 0.5 * 4.905 N = 2.4525 N.
Final Result
The contact force exerted by each wall on the block is approximately 4.905 N, and the frictional force opposing the motion is about 2.4525 N. This analysis shows how the forces interact to keep the block in motion between the walls while considering the effects of friction and gravity.
