Question

An 5kg object is released from rest near the surface of a planet. The vertical position of the object as a function of time is shown in the graph. All frictional forces are considered to be negligible. What is the closest approximation of the weight of the object.
a) 300N
b) 30N
c) 5N
d) 150N

Answer 1

Answer:

The correct option is b: 30 N.

Explanation:

First, we need to find the acceleration due to gravity (a):

$y_{f} - y_{0} = v_{o}t - \frac{1}{2}a(\Delta t)^{2}$   (1)

Where:

$y_{f}$: is the final vertical position (obtained from the graph)

$y_{0}$: is the initial vertical position (obtained from the graph)

v₀: is the initial speed = 0 (it is released from rest)

Δt: is the variation of time (from the graph)

From the graph, we can take the following values of height and time:

t₀ = 0 s → $t_{f}$ = 5 s

y₀ = 300 m → $y_{f}$ = 225 m

Now, by entering the above values into equation (1) and solving for "a" we have:

$a = 2\frac{y_{0} - y_{f}}{(t_{f} - t_{0})^{2}} = 2\frac{300 m - 225 m}{(5 s - 0)^{2}} = 6 m/s^{2}$

Finally, the weight of the object is:

$W = ma = 5 kg*6 m/s^{2} = 30 N$

Therefore, the correct option is b: 30 N.

I hope it helps you!