Take a look at the weather map. The front seen there causes short periods of storms and heavy rains. What type of front is this?
2 answers:
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Answer:
F = 9.81 [N]
Explanation:
To solve this problem we must use Newton's third le which tells us that the sum of forces on a body that remains static must be equal to one resulting from these forces in the opposite direction.
Let's perform a summation of forces on the vertical axis-y to determine the normal force N.
∑F = 0 (axis-y)
where:
m = mass = 4 [kg]
g = gravity acceleration = 9.81 [m/s²]
Now we know that the frictional force can be calculated using the following equation.
f = μ*N
where:
f = friction force [N]
μ = friction coefficient = 0.25
N = normal force = 39.24 [N]
Now replacing:
Then we perform a sum of forces on the X-axis equal to zero. This sum of forces allows us to determine the minimum force to be able to move the object in a horizontal direction.
∑F = 0 (axis-x)
If the coefficient was smaller, a smaller force (F) would be needed to start the movement, this can be easily seen by replacing the value of 0.25, by smaller values, such as 0.1 or 0.05.
If the coefficient were larger, a larger force would be needed.
(a) 907.5 N/m
The force applied to the spring is equal to the weight of the object suspended on it, so:
The spring obeys Hook's law:
where k is the spring constant and is the stretching of the spring. Since we know
, we can re-arrange the equation to find the spring constant:
(b) 1.45 cm
In this second case, the force applied to the spring will be different, since the weight of the new object is different:
So, by applying Hook's law again, we can find the new stretching of the spring (using the value of the spring constant that we found in the previous part):
(c) 3.5 J
The amount of work that must be done to stretch the string by a distance is equal to the elastic potential energy stored by the spring, given by:
Substituting k=907.5 N/m and , we find the amount of work that must be done: