at what inclination should the object be released from rest if it is to reach a maximum horizontal distance of 2m as shown. coeff of friction is 0.37 for all surfaces of contact.
To determine the inclination at which the object should be released from rest in order to reach a maximum horizontal distance of 2m, we need to consider the forces acting on the object and analyze the physics involved.
Let's break down the problem and go through the steps one by one:
1. Identify the given information:
- The object needs to travel a maximum horizontal distance of 2m.
- The coefficient of friction is 0.37 for all surfaces of contact.
2. Determine the relevant forces:
- Weight (mg): This is the force pulling the object vertically downwards. Its magnitude is given by the product of the mass (m) of the object and the acceleration due to gravity (g).
3. Analyze the forces on an inclined plane:
- Decompose the weight force into two components: one parallel to the inclined plane (mg sinθ) and one perpendicular to the inclined plane (mg cosθ), where θ is the angle of inclination.
4. Find the frictional force:
- The frictional force is given by the product of the coefficient of friction (μ) and the perpendicular component of the weight (mg cosθ).
5. Write the net force equation in the horizontal direction:
- The only force acting in the horizontal direction is the frictional force, which opposes the object's motion. Therefore, the net force equation becomes:
Frictional Force = μ * (mg cosθ) = m * a (where a is the acceleration)
6. Determine the acceleration:
- Since the object is released from rest, its initial velocity (v0) is 0. The final velocity (vf) can be calculated using the equation: vf^2 = v0^2 + 2a * d (where d is the distance travelled horizontally, given as 2m in this case).
- Plug in the values: vf^2 = 0 + 2a * 2m = 4a.
- It's important to note that the object will reach its maximum horizontal distance when the acceleration is zero (at the point where it starts to move downhill but hasn't accelerated yet).
- Therefore, at its maximum horizontal distance, vf will also be zero. So the equation becomes: 0^2 = 4a.
- This implies that a = 0.
7. Solve for the angle of inclination:
- Plug in the values of a = 0 and solve the net force equation for θ:
μ * (mg cosθ) = m * a
μ * (mg cosθ) = m * 0
μ * (mg cosθ) = 0
cosθ = 0
- Therefore, the maximum horizontal distance of 2m can be achieved when cosθ is equal to zero.
- Recall that cosθ is equal to zero when θ = 90 degrees (or π/2 radians).
Therefore, the object should be released from rest at an inclination angle of 90 degrees (or π/2 radians) to reach a maximum horizontal distance of 2m, assuming the coefficient of friction remains constant and no other external factors are involved.