A 5.5 kg block is released from rest on a frictionless inclined plane making an angle of 23.3° with the horizontal. Calculate the time it will take this block when released from rest to travel a distance of 0.7 m along the ramp.

sin23.3 = h/0.7

h = 0.7*sin 23.3 = 0.277 m.

h = 0.5g*t^2
h = 0.277 m.
g = 9.8 m/s^2
Solve for t.

To calculate the time it will take for the block to travel a distance of 0.7 m along the ramp, we can use the equations of motion. First, we need to resolve the weight of the block into two components: one parallel to the ramp and one perpendicular to the ramp.

The weight of the block is given by the equation:

Weight = mass x gravity

Where the mass of the block is 5.5 kg and the acceleration due to gravity is approximately 9.8 m/s^2.

Weight = 5.5 kg x 9.8 m/s^2
Weight = 53.9 N

The perpendicular component of the weight (Fn) is given by:

Fn = Weight x cos(angle)

Where the angle is 23.3°.

Fn = 53.9 N x cos(23.3°)
Fn = 48.2 N

The parallel component of the weight (Fp) is given by:

Fp = Weight x sin(angle)

Fp = 53.9 N x sin(23.3°)
Fp = 23.2 N

Now we can calculate the acceleration of the block along the ramp using Newton's second law:

Fp = mass x acceleration

23.2 N = 5.5 kg x acceleration

acceleration = 23.2 N / 5.5 kg
acceleration ≈ 4.22 m/s^2

Next, we can use the kinematic equation to find the time it will take for the block to travel a distance of 0.7 m along the ramp:

distance = initial velocity x time + (1/2) x acceleration x time^2

Since the block is released from rest, the initial velocity is 0. Therefore, the equation simplifies to:

distance = (1/2) x acceleration x time^2

Rearranging the equation, we get:

time^2 = (2 x distance) / acceleration

time^2 = (2 x 0.7 m) / 4.22 m/s^2
time^2 ≈ 0.332 m / (4.22 m/s^2)
time^2 ≈ 0.079 s^2

Finally, we can take the square root of both sides to find the time:

time ≈ √0.079 s^2
time ≈ 0.28 s

Therefore, it will take approximately 0.28 seconds for the block to travel a distance of 0.7 m along the ramp when released from rest.