a 10-kg bos slides down a frictionless ramp with acceleration 3.6 m/s calculate the weight....the angle and normal force
Weight W = M g = 98 N
Acceleration (a) = F/M
where F = W sin A
a = M g sin A/M = g sin A
sinA = 3.6/9.8 = 0.367
A = 21.6 degrees
normal force = W cos 21.6 = ____ N
how did u get A=21.6
To solve this problem, we need to understand the concepts of weight, acceleration, and the forces acting on an object on an inclined plane.
1. Weight: Weight is the force experienced by an object due to the gravitational pull of the Earth. It is given by the formula: Weight = mass × gravitational acceleration.
2. Acceleration: Acceleration is the rate of change of velocity of an object. In this case, the given acceleration is 3.6 m/s.
3. Forces on an inclined plane: When an object is on an inclined plane, there are two main forces acting on it - the weight (mg) acting vertically downward and the normal force (N) acting perpendicular to the surface. The weight can be split into two components: one parallel to the plane (mg sinθ) and one perpendicular to the plane (mg cosθ), where θ is the angle of the incline.
Now, let's calculate the weight, angle, and normal force step by step:
Step 1: Calculate the weight of the object.
Weight = mass × gravitational acceleration
Weight = 10 kg × 9.8 m/s²
Weight = 98 N
The weight of the object is 98 N.
Step 2: Calculate the angle of the incline.
Since the object is sliding down a frictionless ramp, we can use the formula: a = g sinθ, where a is the acceleration and g is the gravitational acceleration.
Rearranging the formula, we get:
sinθ = a / g
sinθ = 3.6 m/s / 9.8 m/s²
sinθ ≈ 0.3673
Taking the inverse sin (arcsin) of both sides, we can find the angle:
θ ≈ arcsin(0.3673)
θ ≈ 21.8°
The angle of the incline is approximately 21.8°.
Step 3: Calculate the normal force.
The normal force (N) is the force exerted by the surface perpendicular to the incline. In this case, it is equal in magnitude but opposite in direction to the perpendicular component of the weight (mg cosθ).
N = mg cosθ
N = 10 kg × 9.8 m/s² × cos(21.8°)
N ≈ 88.1 N
The normal force is approximately 88.1 N.
To summarize:
- The weight of the 10 kg bos is 98 N.
- The angle of the incline is approximately 21.8°.
- The normal force acting on the ramp is approximately 88.1 N.