Question

A 6.0 kg block is released from rest 80 m above
the ground. When it has fallen 60 m its kinetic energy
is approximately (g 10 m/2)
(1) 4800 J
(2) 3600 J
(3) 1200 J
(4) 120 J​

Answer 1

In this type of questions, it is best to apply : CONSERVATION OF MECHANICAL ENERGY principle.

• The change in potential energy will be equal to the change in kinetic energy.

$\Delta KE= - \Delta PE$

$\implies KE - 0= - mg(60 - 80)$

$\implies KE = mg(80 - 60)$

$\implies KE = 6 \times 10 \times (80 - 60)$

$\implies KE = 6 \times 10 \times 20$

$\implies KE = 1200 \: joule$

So, KE of object at that point is 1200 J.

Answer 2

Answer:

3)1200

Explanation:

first we will find v by using this formula

v=√2g(h1-h2) by putting values

v=√2×10×(80-60)

v=20

K.E=1/2mv²

K.E=1/2×6×(20)²

K.E=1/2×6×400

K.E=1200J