Answer to Question #224618 in Mechanical Engineering for Papi Chulo

Question

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

Answer 1

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

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