Question

A projectile of mass 6.8 kg kg is shot horizontally with an initial speed of 14.5 m/s from a height of 26.7 m above a flat desert surface. The acceleration of gravity is 9.81 m/s². For the instant before the projectile hits the surface, find the work done on the projectile by gravity. Answer in units of J.

Answer 1

Answer:

Explanation:

Given

mass of projectile $m=6.8\ kg$

initial horizontal speed $u_x=14.5\ m/s$

height $h=26.7\ m$

Considering vertical motion

velocity gained by projectile during 26.7 m motion

$v^2-u^2=2 as$

v=final velocity

u=initial velocity

a=acceleration

s=displacement

$v^2-(0)^2=2\times (9.8)\times (26.7)$

$v=\sqrt{523.32}$

$v=22.87\ m/s$

Horizontal velocity will remain same as there is no acceleration

final velocity $v_{net}=\sqrt{(v)^2+(u_x)^2}$

$v_{net}=\sqrt{733.57}=27.08\ m/s$

Initial kinetic Energy $K_i=\frac{1}{2}mu_x^2$

$K_i=\frac{1}{2}\times 6.8\times (14.5)^2=714.85\ J$

Final Kinetic Energy $K_f=\frac{1}{2}mv_{net}^2$

$K_f=\frac{1}{2}\times 6.8\times (27.08)^2$

$K_f=2493.30\ J$

Work done by all the force is equal to change in kinetic Energy of object

Work done by gravity is $W_g$

$W_g=\Delta K$

$W_g=2493.30-714.85=1778.45\ J$