3. a 10 kg box is motionless on the floor. if the coefficient of static friction is 0.4 and the coefficient…

3. a 10 kg box is motionless on the floor. if the coefficient of static friction is 0.4 and the coefficient of kinetic friction is 0.3 (between the box and the floor), find the force required to start the block in motion.\n4. it takes a 50 n horizontal force to pull a 20 kg object along the ground at a constant velocity. what is the coefficient of kinetic friction?

3. a 10 kg box is motionless on the floor. if the coefficient of static friction is 0.4 and the coefficient of kinetic friction is 0.3 (between the box and the floor), find the force required to start the block in motion.\n4. it takes a 50 n horizontal force to pull a 20 kg object along the ground at a constant velocity. what is the coefficient of kinetic friction?

Answer

Explanation:

Step1: Analyze the forces acting on the object

When the object is moving at a constant velocity, the horizontal force (F) applied is equal to the force of kinetic friction (F_{k}). The formula for the force of kinetic friction is (F_{k}=\mu_{k}N), where (N = mg) (normal force, (m) is the mass of the object and (g = 9.8\ m/s^{2}) is the acceleration due to gravity).

Step2: Calculate the normal force

Given (m = 20\ kg), then (N=mg=20\times9.8 = 196\ N)

Step3: Solve for the coefficient of kinetic friction (\mu_{k})

Since (F = F_{k}) and (F = 50\ N), from (F_{k}=\mu_{k}N), we can solve for (\mu_{k}). Rearranging the formula gives (\mu_{k}=\frac{F_{k}}{N}). Substituting (F_{k}=50\ N) and (N = 196\ N), we get (\mu_{k}=\frac{50}{196}\approx0.26)

Answer:

The coefficient of kinetic friction is approximately (0.26)