A block lies on a smooth inclined plane tilted at an angle of 35◦ to the horizontal.
a) Draw the free-body diagram for the block.
b) Determine the block’s acceleration as it slides down the inclined plane.
Also
c) If the block started from rest 8.5 m up the incline from its base, determine the block’s speed when it reaches the bottom of the incline.
d) How long did it take the block to reach the bottom of the inclined plane?
a) Well, if I drew a free-body diagram for a block, it might look like a stick figure doing yoga. But in all seriousness, let's focus on the important details. We have our block on an inclined plane, so we need to consider the forces acting on it. We have the weight of the block acting vertically downward, and we also have the normal force acting perpendicular to the plane. Finally, we have the force of friction opposing the motion. So, it's a bit like a tug-of-war with these forces.
b) As for the block's acceleration, we would need some more information. Are we talking about a block sliding down the plane without any external forces? Is there some coefficient of friction involved? If you can provide those details, I'll be able to give you a more clownish answer!
a) To draw the free-body diagram for the block, we need to consider the forces acting on it. The only forces acting on the block are:
1. The weight of the block (mg), which acts vertically downwards.
2. The normal force (N) exerted by the inclined plane, which acts perpendicular to the surface of the plane.
3. The frictional force (f) acting parallel to the plane.
The arrow representing each force should be drawn in the direction in which it acts. The free-body diagram will look like this:
------
| |
| mg |
| |
|-------|
|
| N
|
|------------------> Direction of motion
b) To determine the block's acceleration as it slides down the inclined plane, we can use the following equation derived from Newton's second law:
Net force = mass × acceleration
The net force acting on the block is the component of the weight (mg) parallel to the inclined plane minus the frictional force (f). We can express the weight component as mg × sin(35°) and the frictional force as f = μN.
Applying Newton's second law to the forces acting along the plane, we have:
Net force = m × a
(mg × sin(35°)) - (μN) = m × a
Now, the normal force (N) can be calculated as N = mg × cos(35°).
Substituting the value of N into the equation, we get:
(mg × sin(35°)) - (μ(mg × cos(35°))) = m × a
Simplifying the equation, we have:
mg(sin(35°) - μcos(35°)) = m × a
Dividing both sides by m, we get:
g(sin(35°) - μcos(35°)) = a
Now we can substitute the values:
g(sin(35°) - μcos(35°)) = a
a ≈ g(sin(35°) - μcos(35°))
where g is the acceleration due to gravity (approximately 9.8 m/s^2) and μ is the coefficient of friction between the block and the inclined plane.
a) To draw the free-body diagram for the block, we need to consider all the forces acting on it.
1. The force of gravity (mg): This force acts vertically downwards and can be represented by a downward arrow. Its magnitude is given by mg, where m is the mass of the block and g is the acceleration due to gravity.
2. The normal force (N): This force acts perpendicular to the inclined plane and prevents the block from sinking into the plane. Since the plane is smooth, the normal force is equal in magnitude and opposite in direction to the component of the gravitational force acting perpendicular to the plane. So, N can be represented by an arrow pointing upwards.
3. The force along the inclined plane (F): This force is parallel to the inclined plane and causes the block to slide down. It is the force we need to determine in part b. We represent it with an arrow pointing down the inclined plane.
b) To determine the block's acceleration as it slides down the inclined plane, we need to consider the forces acting on the block along the inclined plane.
1. The force of gravity can be resolved into two components: one perpendicular to the plane (mg cosθ) and one parallel to the plane (mg sinθ), where θ is the angle of the inclined plane.
2. The force along the inclined plane (F) is the force that causes the block to slide down. There is no friction in this case since the plane is smooth, so F is the only force acting in the direction of motion.
Using Newton's second law, we can write the equation of motion for the block along the inclined plane:
F = m・a
F = mg sinθ
Since the force along the inclined plane is equal to mg sinθ, we can substitute it into the equation of motion:
mg sinθ = m・a
Canceling out the mass 'm' from both sides of the equation gives us:
g sinθ = a
So, the block's acceleration as it slides down the inclined plane is equal to g multiplied by sinθ, where g is the acceleration due to gravity and θ is the angle of the inclined plane.
(b) a = g sin35°
(c) v^2 = 2as
(d) s = 1/2 at^2