Question

Erfan pushes a 5kg crate of stage props starting from rest along the smooth and slippery horizontal floor of a theater stage. The crate moves with a constant acceleration for 10 seconds as Erfan pushes with a force of 0.22 N to reach the edge of the stage. Then the crate runs off the edge of the stage and lands in the orchestra pit below. The floor of the orchestra pit is 3 meters below the edge of the stage. Assume friction is negligible. How far away from the stage edge does the crate land?

Answer 1

distance S = 0.3443 m

Step-by-step explanation:

given data

mass m = 5 kg

constant acceleration a = 10 seconds

force f = 0.22 N

distance = 3 meter

solution

we will apply Newton's second law

and we find here acceleration of the crate that is

acceleration a =
`(forca)/(mass)` ............1

acceleration a =
`(0.22)/(5)`

acceleration a = 0.044 m/s²

now we get here velocity of crate just before running off edge

v = u + at ...........2

v = 0 + 0.044 × 10

velocity v = 0.44 m/s

and

now we get here time taken for that flight is as

h = 0.5 × gt² ..............3

3 = 0.5 × 9.8× t²

time t = 0.7825 seconds

and horizontal distance will be here

distance S = vt

distance S = 0.44 × 0.7825

distance S = 0.3443 m