A body of mass 2 kg is dropped from the top of a tower of height 100m. If the acceleration due to gravity is 10 m/s2 calculate the K.E at the end of 5 seconds
To calculate the kinetic energy (K.E.) of the body at the end of 5 seconds, we need to determine its final velocity, which will be achieved by using the equations of motion.
First, let's find the final velocity of the body using the equation:
v = u + at
where:
v = final velocity of the body
u = initial velocity of the body (which is 0 m/s as the body is dropped from rest)
a = acceleration due to gravity (which is 10 m/s^2)
t = time taken (which is 5 seconds)
Substituting the given values:
v = 0 + (10 m/s^2 × 5 s)
v = 50 m/s
Now that we have the final velocity, we can calculate the kinetic energy of the body using the equation:
K.E. = 1/2 × m × v^2
where:
K.E. = kinetic energy
m = mass of the body (which is 2 kg)
v = final velocity of the body (which is 50 m/s)
Substituting the given values:
K.E. = 1/2 × 2 kg × (50 m/s)^2
K.E. = 1/2 × 2 kg × 2500 m^2/s^2
K.E. = 2500 J
Therefore, the kinetic energy of the body at the end of 5 seconds is 2500 Joules (J).
Note that the body will hit the ground after 4.47 s. The KE after 5 seconds will depend how well it bounces.
trick question.
2500j
v=u+gt
V= 0 + 10*5
V= 50
K.E= 1/2 *2*50*50
K.E = 2500J