Question

A 60-kg crate is lowered from a loading dock to the floor using a top passing over fixed support. The rope exerts a constant upward force on the crate of 500 N.
A. Will the crate accelerate? Explain.
B. What are the magnitude and direction of the acceleration of the crate?
C. How long will it take for the crate to reach the floor of the height of the loading dock is 1.4m above the floor?
D. How fast is the crate traveling when it hits the floor?

Answer 1

Explanation:

a) What is the weight of the crate?

W = m*g = 60 kg *9.81 m/s^2 = 588.6 N

Sum of the forces in the vertical direction on the box (up is +) = F - W = 500 N - 588.6 N = -88.6 N

Since the forces do not add up to 0, then there is a net force on the crate which will cause the crate to accelerate.

b) F = m*a ==> a = F/m

a = -88.6 N / 60 kg = -1.48 m/s^2 (- means down since above we declared up as +)

c) Assume it starts from rest, vi = 0 m/s

Equation of motion in the vertical direction

h = h0 + v0*t + 1/2*a*t^2

h0 = 1.4 m

h = 0 m

v0 = 0 m/s

a = -1.48 m/s^2

0 = 1.4 m + 0 - 1/2* 1.48 m/s^2 * t^2

1.4 m = 1/2 * 1.48 m/s^2 * t^2

t = 1.38 s

d) vf = vi + a*t

vi = 0 m/s

vf = a*t = -1.48 m/s^2 * 1.38 s = -2.04 m/s