A force sensor is used to measure a force applied to a 7.5 kg box on a hard, horizontal surface. It records that the box starts to move when the force is equal to 16.5 N. This force makes the box start to move with an acceleration of 0.5 m/s2. Find s and k respectively.
I keep getting .0510 or -.0510 but it's wrong.
Ff=M*N Ff=Fnet=m*a N=m*g these are formula's i've used..
The force that makes it start moving and accelerating satisfies these two equations:
F - M g k = M a
F = M g s
Therefore
M g (s-k) = M a
a/g = 0.051 = s - k
You can also use the second equation to write
16.5 = M g s = 7.5*9.8 s = 73.5 s
s = 16.5/73.5 = 0.224
k = s - .051 = 0.173
Those are the static and kinetic coefficients of friction, respectively.
To find the values of s and k, we can use Newton's second law of motion, which states that the net force acting on an object is equal to its mass multiplied by its acceleration (Fnet = m * a).
In this case, the force sensor recorded that the box starts to move when the force applied is 16.5 N, and the acceleration of the box is 0.5 m/s^2.
First, let's find the net force acting on the box:
Fnet = m * a
Fnet = 7.5 kg * 0.5 m/s^2
Fnet = 3.75 N
The net force is equal to the force of friction (Ff) acting on the box. The force of friction can be calculated using the equation:
Ff = μ * N
where μ is the coefficient of friction and N is the normal force. In this case, since the box is on a hard, horizontal surface, the normal force is equal to the weight of the box, given by:
N = m * g
N = 7.5 kg * 9.8 m/s^2
N = 73.5 N
Now we can substitute the value of N into the equation for Ff:
Ff = μ * N
3.75 N = μ * 73.5 N
To find the coefficient of friction (μ), divide both sides of the equation by 73.5 N:
μ = 3.75 N / 73.5 N
μ ≈ 0.051
So the coefficient of friction (k) is approximately 0.051.
Finally, to find the value of s, we can use the equation for friction force:
Ff = μ * N
Ff = μ * m * g
Substituting the known values:
Ff = 0.051 * 7.5 kg * 9.8 m/s^2
Ff ≈ 3.55 N
Therefore, the value of s is approximately 3.55 meters.