Question

The figure shows the speed of a person's body as he does a chin-up. Assume the motion is vertical and the mass of the person's body is 64.0 kg. determine the force exerted by the chin-up bar on his body at t = 0s; t = 0.5s; t = 1.1s; and t = 1.6s

Answer 1

Explanation :

It is given that:

Mass of person, m = 64 kg

We know that, F = m a

Where, a is acceleration due to gravity and $a =\dfrac{v}{t}$

(i) Force exerted at t = 0 s :

$\because F = ma$

At t = 0 s, the body is at rest so no acceleration. The force will be zero.

(ii) At t = 0.5 s, force of gravity will also acts i.e. f = m (a+g)

From the graph it is clear that at t = 0.5 s, velocity is 15 cm/s =0.15 m/s

So,$F=64\ kg\times (\dfrac{0.15\ cm/s}{0.5\ s} +9.8\ m/s^2)$

$F = 646.4\ N$

(ii) Force at t = 1.1 s :

At t = 1.1 s, v = 28 cm/s = 0.28 m/s

$F=64\ kg\times (\dfrac{0.28\ m/s}{1.1\ s}+9.8\ m/s^2)$

$F =643.2\ N$

(iii) Force at t = 1.6 s :

At t = 1.6 s, v = 6 cm/s = 0.06 m/s

$F=64\ kg\times (\dfrac{-0.06 \ m/s}{1.6\ s}+9.8\ m/s^2)$ (decelerating)

$F =-627.1\ N$

Hence, this is the required solution .

Answer 2

Please check calculations and let me know later on