Question

1. a 50-kg crate is attached to an electric hoist in a warehouse. the hoist begins lifting the crate with an average upward velocity of 4.0 m/s and maintains a constant upward acceleration of 1 m/s². approximately how long will it take for the hoist to lift the crate to a shelf 10.0 meters above its starting position?

Explanation:

Step1: Identify known values

$d = 10.0\ \text{m}$, $v_0 = 4.0\ \text{m/s}$, $a = 1\ \text{m/s}^2$

Step2: Use kinematic equation

$$d = v_0 t + \frac{1}{2} a t^2$$
Substitute values:
$$10 = 4t + \frac{1}{2}(1)t^2$$
Rearrange to standard quadratic form:
$$0.5t^2 + 4t - 10 = 0$$
Multiply by 2 to simplify:
$$t^2 + 8t - 20 = 0$$

Step3: Solve quadratic equation

Use quadratic formula $t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, where $a=1$, $b=8$, $c=-20$:
$$t = \frac{-8 \pm \sqrt{8^2 - 4(1)(-20)}}{2(1)}$$
$$t = \frac{-8 \pm \sqrt{64 + 80}}{2}$$
$$t = \frac{-8 \pm \sqrt{144}}{2}$$
$$t = \frac{-8 \pm 12}{2}$$
Take positive root (time cannot be negative):
$$t = \frac{-8 + 12}{2} = \frac{4}{2} = 2$$

Answer:

$\boldsymbol{2\ \text{seconds}}$