the small 2kg block A slides down the curve path and passes the lowest point B with a speed 4 m/s. If the radius of the curvature path at B is 1.5 m, determine the normal force N exerted on the block by path at this point. Is knowledge of the friction properties necessary? solution of this question
normal force will be mg+ centripetal or
mg+mv^2/r
31.6
To determine the normal force N exerted on the block by the path at the lowest point B, we can use the concept of centripetal force.
First, let's calculate the acceleration of the block at point B using the formula for centripetal acceleration:
a = v^2 / r
where v is the velocity at point B and r is the radius of curvature at point B.
Given:
v = 4 m/s
r = 1.5 m
Substituting these values into the formula, we have:
a = (4 m/s)^2 / 1.5 m
= 16 m^2/s^2 / 1.5 m
= 10.67 m/s^2
Now, we know that the net force acting on the block at point B is the difference between the gravitational force (mg) and the normal force (N).
The gravitational force (mg) can be calculated using the mass (m) of the block and the acceleration due to gravity (g).
Given:
m = 2 kg
g = 9.8 m/s^2
mg = (2 kg) * (9.8 m/s^2)
= 19.6 N
Therefore, the net force at point B is:
F_net = mg - N
Since the block is moving on a curved path, the net force is equal to the centripetal force (F_c):
F_c = F_net
F_c = ma
ma = mg - N
Solving for N, we have:
N = mg - ma
Substituting the given values, we get:
N = (2 kg) * (9.8 m/s^2) - (2 kg) * (10.67 m/s^2)
= 19.6 N - 21.34 N
≈ -1.74 N
The negative sign indicates that the normal force is directed in the opposite direction of the gravitational force. In this case, it means that the path exerts an upward force on the block.
So, the normal force exerted on the block by the path at point B is approximately 1.74 N.
To determine the normal force exerted on the block by the path at point B, we need to consider the forces acting on the block at that point.
First, let's analyze the forces involved:
1. Gravitational force (weight): The weight of the block (mg) acts vertically downward.
2. Normal force (N): The normal force acts perpendicular to the surface of the path and prevents the block from sinking into the path.
3. Frictional force (f): The frictional force acts parallel to the surface of the path. However, in this question, we are not given any information about friction properties. Therefore, we can assume that the frictional force is not present or negligible.
At point B, the block has passed the lowest point and is moving upward. The speed of the block at point B is 4 m/s. Since there is no friction, the normal force and gravitational force are the only factors influencing the block's motion.
To find the normal force (N), we need to equate the net force acting on the block with the product of its mass (m) and acceleration (a) in the normal direction.
Using Newton's second law, F_net = ma, and considering only the forces acting in the vertical direction, we have:
F_net = N - mg = ma
Since the block is moving along a curved path, it is subjected to a centripetal acceleration (a_c) towards the center of the path. The centripetal acceleration is given by:
a_c = (v^2) / r
Where v is the velocity (4 m/s) and r is the radius of curvature of the path at point B (1.5 m).
Now substituting the centripetal acceleration in the equation for F_net:
N - mg = (v^2) / r
Substituting the given values:
N - (2 kg * 9.8 m/s^2) = (4 m/s)^2 / 1.5 m
N - 19.6 N = 16/1.5 N
N = 19.6 N + 10.67 N ≈ 30.27 N
Therefore, the normal force exerted on the block by the path at point B is approximately 30.27 N.