what is the value for k if a spring stretches 0.4m when submitted to a 1800N force?
How much force is needed to keep a 23 kg box moving at a constant velocity across a warehouse floor if the coefficient of friction between the box and the floor is 0.88?
Both of these are straightforward uses of standard formulas. I will be happy to critique your thinking.
i'm not very good at physics
for the first one do i use the formula F=-Kx to make the answer -4500 ?
and for the second i really have no idea what formula to use
To calculate the value of k in the first question, we need to use Hooke's Law, which states that the force exerted by a spring is proportional to the displacement of the spring from its equilibrium position. Mathematically, it can be represented as F = kx, where F is the force, k is the spring constant, and x is the displacement.
In this case, we are given that the spring stretches 0.4m when subjected to a force of 1800N. We can use this information to calculate the value for k.
To find k, we rearrange the equation: k = F / x.
Substituting the given values, we have: k = 1800N / 0.4m = 4500 N/m.
Therefore, the value of k is 4500 N/m.
For the second question, we can use the equation for frictional force:
Frictional Force = coefficient of friction * normal force.
The normal force is the force exerted by the surface perpendicular to the object, which in this case is equal to the weight of the box.
The formula for weight is weight = mass * acceleration due to gravity, where the acceleration due to gravity is usually approximated as 9.8 m/s².
Given that the mass of the box is 23 kg and the coefficient of friction is 0.88, we can calculate the force required to keep it moving at a constant velocity.
First, find the normal force: Normal Force = mass * gravity = 23 kg * 9.8 m/s² = 225.4 N.
Now, find the frictional force by multiplying the coefficient of friction by the normal force: Frictional Force = 0.88 * 225.4 N = 198.35 N.
Therefore, the force needed to keep the box moving at a constant velocity across the warehouse floor is approximately 198.35 N.