An object of mass 1.5 kg is placed on a rough horizontal surface, as shown in Figure 1. Force F acts on the object. When the magnitude of F is 10 N, the object just begins to slide. Determine the coefficient of static friction between the object and the surface.
i do not know how to find acceleration.
assuming the force is parallel to the surface.
10N=mu*weight
solve for mu.
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To find the coefficient of static friction between the object and the surface, we first need to determine the acceleration of the object when it just begins to slide.
Here's how you can find the acceleration:
1. Begin by using Newton's second law of motion, which states that the force acting on an object (F) is equal to the mass of the object (m) multiplied by its acceleration (a), i.e., F = m × a.
2. In this case, the force acting on the object is the force F applied to it. So, we have F = 10 N.
3. The mass of the object is given as 1.5 kg. Hence, we have m = 1.5 kg.
4. Rearrange the equation to solve for acceleration: a = F / m.
5. Plug in the values: a = 10 N / 1.5 kg = 6.67 m/s².
Now that we have the acceleration, we can proceed to find the coefficient of static friction.
To determine the coefficient of static friction, we can use the formula:
μs = Ff / N,
where μs is the coefficient of static friction, Ff is the force of static friction, and N is the normal force.
In this case, since the object is not moving vertically, the normal force N is equal to the weight of the object, which is given by the formula:
N = m × g,
where g is the acceleration due to gravity, which is approximately 9.8 m/s².
Substituting the given values, we have:
N = 1.5 kg × 9.8 m/s² = 14.7 N.
Now, we can find the force of static friction (Ff) using the formula:
Ff = μs × N.
Since the object is on the verge of sliding, the force of static friction Ff is equal to the applied force F, which is 10 N.
Now we can solve for the coefficient of static friction:
10 N = μs × 14.7 N,
μs = 10 N / 14.7 N = 0.68.
Therefore, the coefficient of static friction between the object and the surface is 0.68.
Remember that this is just an example, and the values may vary depending on the specific problem.