Answer to Question #138199 in Molecular Physics | Thermodynamics for Johannes Steven

Question

1. (i) The palm fruit has a velocity of 11m/s vertically downwards. When it hits the ground it comes to rest in 0.2 s. Calculate the average deceleration on the fruit.

(ii) Calculate the kinetic energy of the palm fruit just before it hits the ground.

Answer 1

Solution

Velocity of palm fruit (v)=11 m/s

Time in which fruit came to rest(t)=0.2sec

So by the first equation of motion

$v=u-a_{avg}t$

Final velocity v=0m/s

u=11 m/s ; and t=0.2sec

So

$a_{avg}=\frac{u}{t}$

$a_{avg}=\frac{11}{0.2}=55 m/s^2$

So average deceleration is 55m/s^2

(B) kinetic energy before strike to the ground can be given as

$K=\frac{1}{2}mv^2$

$K=\frac{1}{2}m(11)^2$

m= mass of palm fruit

$K=60.5m\space J$

So kinetic energy of fruit before striking to ground is 60.5m J