A block slides down a frictionless plane having an inclination of θ = 13.2°. The block starts from rest at the top, and the length of the incline is 1.70 m.

-> Find the acceleration of the block.
(m/s2 down the incline)

-> Find its speed when it reaches the bottom of the incline.
m/s)

weight ccomponent down slope

F = m g sin 13.2
so
m g sin 13.2 = m a
so
a = 9.81 sin 13.2
( what else is new?)

d = (1/2) a t^2
1.7 = (1/2) a t^2
solve for t
then
v = a t

Well, if I were a block sliding down a frictionless plane, I would definitely pick up some speed! But let me calculate this for you:

To find the acceleration of the block, we can use good old Newton's second law, F = ma. In this case, the only force acting on the block is the component of its weight parallel to the incline, which we can calculate as mg * sin(θ), where m is the mass of the block and g is the acceleration due to gravity.

So, the acceleration (a) of the block can be calculated using the equation:

a = (mg * sin(θ)) / m

Since the mass of the block cancels out, the acceleration is simply g * sin(θ). And g, being the acceleration due to gravity, is approximately 9.8 m/s^2. So, the acceleration of the block down the incline is about 9.8 m/s^2 * sin(13.2°).

Now, to find the speed of the block when it reaches the bottom of the incline, we can use the kinematic equation:

v^2 = u^2 + 2as

Where v is the final velocity, u is the initial velocity (which is 0 in this case since the block starts from rest), a is the acceleration we just calculated, and s is the distance traveled down the incline (1.70 m).

Plugging in the values, we get:

v^2 = 0 + 2 * (9.8 m/s^2 * sin(13.2°)) * 1.70 m

After simplifying the equation, we find that the speed (v) of the block when it reaches the bottom of the incline is approximately the square root of that expression we just calculated.

I hope the block enjoys its speedy descent down the incline!

To find the acceleration of the block as it slides down the incline, we can use the equation:

acceleration = g * sin(θ)

Where g is the acceleration due to gravity (approximately 9.8 m/s^2) and θ is the angle of inclination (13.2°).

acceleration = 9.8 * sin(13.2)

Using a scientific calculator, we can find:

acceleration ≈ 2.165 m/s^2

So, the acceleration of the block down the incline is approximately 2.165 m/s^2.

To find the speed of the block when it reaches the bottom of the incline, we can use the equation:

velocity = √(2 * acceleration * distance)

Where acceleration is the value we just calculated and distance is the length of the incline (1.70 m).

velocity = √(2 * 2.165 * 1.70)

Using a scientific calculator, we can find:

velocity ≈ 3.774 m/s

So, the speed of the block when it reaches the bottom of the incline is approximately 3.774 m/s.

To find the acceleration of the block sliding down the frictionless plane, we can use the equation of motion for an object on an inclined plane:

a = g * sin(θ)

where:
a = acceleration of the block
g = acceleration due to gravity, approximately 9.8 m/s^2
θ = angle of inclination of the plane, which is given as 13.2°

Now let's substitute the values into the equation and calculate the acceleration:

a = 9.8 * sin(13.2°)
a ≈ 9.8 * 0.228
a ≈ 2.24 m/s^2

So, the acceleration of the block sliding down the incline is approximately 2.24 m/s^2 down the incline.

To find the speed of the block when it reaches the bottom of the incline, we can use the equation:

v = sqrt(2 * a * d)

where:
v = final velocity of the block
a = acceleration of the block (which we calculated as 2.24 m/s^2)
d = distance traveled along the incline, which is given as 1.70 m

Now let's substitute the values into the equation and calculate the final velocity:

v = sqrt(2 * 2.24 * 1.70)
v ≈ sqrt(7.59)
v ≈ 2.76 m/s

Therefore, the speed of the block when it reaches the bottom of the incline is approximately 2.76 m/s.