Question

A crate slides from rest and accelerates uniformly at 4.9m/s^2 along a frictionless roof 3m long which is inclined at an angle of 30 degrees to the horizontal. Determine (i)the velocity of the crate just after losing contact with the roof (ii) the velocity(magnitude and direction just before it hits the ground (iii)the time the crate takes to hit the ground after leaving the roof (iiii) the horizontal distance between the point directly below the roof and the landing point

Answer 1

Step-by-step explanation:

This is quite simple, we can solve this problem using kinematic equations of motion.

(i) The velocity of the crate just after losing contact with the roof:

We can use the kinematic equation, v^2 = u^2 + 2as, where u = 0 m/s (initial velocity), a = 4.9 m/s^2 (acceleration), and s = 3 m (distance).

v^2 = 0^2 + 2(4.9)(3) = 29.4

v = √29.4 = 5.42 m/s (velocity just after losing contact with the roof)

(ii) The velocity (magnitude and direction) just before it hits the ground:

We can use the kinematic equations of motion for motion in a straight line with constant acceleration.

First, let's find the time taken by the crate to reach the ground after leaving the roof. We can use the kinematic equation, s = ut + 1/2 at^2, where s = 3 m (vertical distance), u = 5.42 m/s (initial velocity), a = 9.8 m/s^2 (acceleration due to gravity), and t is the time taken to reach the ground.

3 = 5.42t - 1/2 (9.8)t^2

Simplifying this quadratic equation, we get t = 0.88 s (time taken to reach the ground after leaving the roof).

Now, let's find the velocity just before it hits the ground. We can use the kinematic equation, v = u + at, where u = 5.42 m/s (initial velocity), a = 9.8 m/s^2 (acceleration due to gravity), and t = 0.88 s (time taken to reach the ground after leaving the roof).

v = 5.42 + (9.8)(0.88) = 14.74 m/s (magnitude of the velocity just before hitting the ground)

The direction of the velocity just before hitting the ground is downwards, at an angle of 30 degrees below the horizontal (due to the slope of the roof).

(iii) The time the crate takes to hit the ground after leaving the roof:

We have already calculated this in part (ii), which is t = 0.88 s.

(iv) The horizontal distance between the point directly below the roof and the landing point:

We can use the formula, distance = velocity × time.

The horizontal velocity of the crate is constant and equal to 5.42 m/s, which is the same as the velocity just after losing contact with the roof (since there is no friction).

Therefore, the horizontal distance traveled by the crate before hitting the ground is:

distance = 5.42 m/s × 0.88 s = 4.77 m.

Answer 2

(i) To find the velocity of the crate just after losing contact with the roof, we can use the equation:

v^2 = u^2 + 2as

where v is the final velocity, u is the initial velocity (which is zero), a is the acceleration, and s is the displacement (which is 3 m).

Substituting the values, we get:

v^2 = 0 + 2 × 4.9 × 3

v^2 = 29.4

v = √29.4

v ≈ 5.42 m/s

Therefore, the velocity of the crate just after losing contact with the roof is approximately 5.42 m/s.

(ii) To find the velocity just before the crate hits the ground, we can use the conservation of energy principle. The initial potential energy of the crate at the top of the roof is converted into kinetic energy just before it hits the ground.

Initial potential energy = mgh

where m is the mass of the crate, g is the acceleration due to gravity, and h is the height of the roof.

Final kinetic energy = (1/2)mv^2

where v is the velocity just before the crate hits the ground.

Since there is no loss of energy, we can equate these two values:

mgh = (1/2)mv^2

Canceling out m on both sides and substituting the values, we get:

(1/2) × 1.17 × 9.81 × 3 = (1/2) × 1.17 × v^2

v^2 = 34.5

v = √34.5

v ≈ 5.87 m/s

Therefore, the velocity just before the crate hits the ground is approximately 5.87 m/s, and its direction is along the horizontal.

(iii) To find the time taken by the crate to hit the ground after leaving the roof, we can use the equation:

s = ut + (1/2)at^2

where s is the displacement (which is the height of the roof, 3 m), u is the initial velocity (which is zero), a is the acceleration, and t is the time taken.

Substituting the values, we get:

3 = 0 + (1/2) × 9.81 × t^2

t^2 = 0.612

t ≈ √0.612

t ≈ 0.78 s

Therefore, the time taken by the crate to hit the ground after leaving the roof is approximately 0.78 s.

(iv) Finally, to find the horizontal distance between the point directly below the roof and the landing point, we can use the equation:

s = ut + (1/2)at^2

where s is the horizontal distance, u is the initial velocity (which is zero), a is the acceleration (which is zero, since there is no force acting along the horizontal direction), and t is the time taken (which is 0.78 s).

Substituting the values, we get:

s = 0 + (1/2) × 0 × (0.78)^2

s = 0

Therefore, the horizontal distance between the point directly below the roof and the landing point is zero, which means the crate lands directly below the roof.