Question

A pole-vaulter just clears the bar at 5.31 m and falls back to the ground. The change in the vaulter's potential energy during the fall is -3900 J. What is his weight?

Answer 1

Answer:

Weight = 734.46 N

Explanation:

Given:

Initial height of the pole-vaulter is, $h_i=5.31\ m$

Final height of the pole-vaulter is, $h_f=0\ m$

Change in the vaulter's potential energy is, $\Delta U=-3900\ J$

We know that, change in potential energy is given as:

$\Delta U=mg(h_f-h_i)$

Where, 'm' is the mass of the object, 'g' is acceleration due to gravity and has a value of 9.8 m/s².

Now, weight of the object is given as the product of its mass and acceleration due to gravity. So, replace 'mg' by weight 'w'. So, the equation becomes,

$\Delta U = w(h_f-h_i)$

Now, rewriting in terms of 'w', we get:

$w=\dfrac{\Delta U}{(h_f-h_i)}$

Now, plug in all the given values and solve for 'w'. This gives,

$w=\dfrac{-3900\ J}{(0-5.31)\ m}\\\\\\w=\dfrac{3900}{5.31}\ N\\\\\\w=734.46\ N$

Therefore, weight of the vaulter is 734.46 N.