box of books weighs 350 n is shoved across the floor with a force of 500 n directed downward at angle of 37 with horizontal. if muk between box and floor is .5 how long does it take box to move 4 m starting from rest and what is largest value of muk that will allow the box to move with that applied forcep
500 cos 37 - .5 (350 + 500 sin 37) = m a
m = 350/9.81
find a
4 = (1/2) a t^2
for part be find mu for:
500 cos 37 - mu (350 + 500 sin 37) = 0
To answer your question, we need to first consider the forces acting on the box.
1. Weight (W): The weight of the box is given as 350 N and is directed downward vertically.
2. Applied force (F): The applied force is given as 500 N directed downward at an angle of 37 degrees with the horizontal.
3. Normal force (N): The normal force exerted by the floor on the box is equal in magnitude and opposite in direction to the weight of the box.
4. Frictional force (f): The frictional force opposing the motion is responsible for the box's acceleration.
The frictional force is given by the equation: f = μk * N, where μk is the coefficient of kinetic friction and N is the normal force.
In this case, the normal force N is equal to the weight W since the box is resting on a horizontal surface. Therefore, N = W = 350 N.
Now, let's find the coefficient of kinetic friction (μk). We are given μk = 0.5.
So, f = 0.5 * 350 N = 175 N.
Since the applied force is greater than the maximum possible friction force, the box will move.
To find the acceleration of the box, we can use Newton's second law: F - f = m * a, where F is the applied force, f is the frictional force, m is the mass of the box, and a is the acceleration. Rearranging the equation gives us a = (F - f) / m.
The mass (m) can be found using the formula: W = m * g, where W is the weight and g is the acceleration due to gravity (9.8 m/s²). Rearranging the equation gives us m = W / g.
m = 350 N / 9.8 m/s² ≈ 35.71 kg.
Now, we can calculate the acceleration:
a = (500 N - 175 N) / 35.71 kg ≈ 9.86 m/s².
The box starts from rest and moves a distance of 4 m. We can use the equation of motion: s = ut + (1/2)at², where s is the distance, u is the initial velocity (which is 0 in this case), a is the acceleration, and t is the time.
Since we want to find the time, rearranging the equation gives us t = sqrt((2s) / a), where sqrt refers to the square root.
t = sqrt((2 * 4 m) / 9.86 m/s²) ≈ 0.899 s.
Therefore, it will take approximately 0.899 seconds for the box to move 4m starting from rest, given the conditions described.
As for the largest value of μk that allows the box to move with the applied force, we can calculate this by finding the maximum frictional force that can be applied without exceeding the given applied force.
Since F = f for maximum static friction, we can substitute F = μs * N into the equation. Here, μs represents the coefficient of static friction. Therefore, μs * N = F.
μs * N = 500 N.
Since N = W = 350 N:
μs * 350 N = 500 N.
μs = 500 N / 350 N = 1.43.
Hence, the largest value of the coefficient of static friction (μs) that will allow the box to move with the given applied force is approximately 1.43.