103 469
Assignments Done
98.8%
Successfully Done
In November 2021

# Answer to Question #224618 in Mechanical Engineering for Papi Chulo

Question #224618

An elevator whose floor to ceiling height is 2.75m starts ascending with a constant upward acceleration of 1.2 m/s2. Two seconds after the start, a loose bolt drops from the ceiling towards the floor of the elevator. Calculate:

a)The time taken by the bolt to reach the floor

b)Displacement of the bolt during that time

c)The distance travelled by the bolt during the same time

1
2021-08-11T07:28:42-0400

Part a

Relative velocity is zero

s= ut +\frac{1}{2}at^2\\ 2.75= 0*t +\frac{1}{2}*11*t^2\\ \implies t = 0.7 s

Part b

\Delta y = V_0 t+ \frac{1}{2}gt^2\\ \Delta y = 2.4*0.7+ \frac{1}{2}*9.81*0.7^2\\ \Delta y = -0.7 m

Part c

v^2=u^2-2gs\\ s= \frac{v^2}{2g}\\ s= \frac{2.4^2}{2*9.81}\\ s= 0.29 m\\ Total = 2*0.29 +0.7 \\ Total = 1.3 m

Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!