A box of mass m = 0.20 kg slides across frictionless tabletop from rest, with 2.5 m/s2 acceleration.
The length of the tabletop is L = 2.00 m and its height is H = 1.50 m. The box then slides off the edge and hits the floor at some distance away, Find the distance x.
v=sqrt(2as) =sqrt(2•2.5•2)=3.16 m/s,
H=gt²/2 => t =sqrt(2H/g) =
=sqrt(2•1.5/9.8) =0.55 s.
X =v•t =3.16•0.55 = 1.75 m.
To find the distance x that the box travels after sliding off the edge and hitting the floor, we can use the equations of motion for a horizontally launched projectile. The box will have an initial velocity in the horizontal direction and a constant acceleration in the vertical direction due to gravity.
Let's break down the problem into two parts: the motion on the tabletop and the motion after sliding off the edge.
1. Motion on the tabletop:
We are given the acceleration on the tabletop, a = 2.5 m/s^2. We can use the equation of motion:
v^2 = u^2 + 2as
where v is the final velocity, u is the initial velocity, a is the acceleration, and s is the distance travelled.
Since the box starts from rest on the tabletop, u = 0 m/s. The final velocity can be found using the equation with v^2 as the unknown:
v^2 = 0^2 + 2 * 2.5 * 2.00
v^2 = 10
v = √10 m/s
Now we know the horizontal velocity and can move on to the next part.
2. Motion after sliding off the edge:
Once the box slides off the edge, it becomes a projectile moving in both the horizontal and vertical directions simultaneously. We are interested in finding the horizontal distance x traveled before hitting the ground.
We need to find the time of flight (time in the air) and then use it to calculate the horizontal distance traveled.
a) Time of flight:
The formula for the time of flight is:
t = 2 * u * sin(theta) / g
where u is the initial velocity in the horizontal direction, theta is the angle of launch (which is zero since the box is sliding horizontally), and g is the acceleration due to gravity (approximately 9.8 m/s^2).
In this case, the horizontal velocity (u) is given by v from the previous part: u = √10 m/s.
Hence:
t = 2 * √10 * sin(0) / 9.8
t = 0 / 9.8
t = 0 s
b) Horizontal distance traveled:
Using the formula for horizontal distance:
x = u * t
x = √10 * 0
x = 0 m
Hence, the box travels a horizontal distance of 0 m after sliding off the edge and hitting the floor.