Question

An object of mass 70 kg is raised to a height of 10m above the ground, the potential energy is equal to (g= 10m/s^2)*​

Answer 1

Answer:

It is given that,

Mass of the object, m = 70 kg

Height above which it is taken, h = 10 m

The potential energy of the particle is said to be possessed by virtue of its position. It is given by :

P = 48020 J

When the object is half way down i.e. d = 5 m

let v is the velocity of the object. It can be calculated using third equation of motion as :

v = 9.89  m/s

So, when the object is allow to fall, let K is its kinetic energy. It is given by :

K = 3423.42 J

So, the kinetic energy of the object when it is allowed to fall when it is half way down is 3423.42 J. Hence, this is the required solution.

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Kinetic energy

Explanation:

Answer 2

Answer:

P.E (potential energy) = 7000 J. K.E (kinetic energy) = 3500 J

Explanation:

because  

P.E = mgh

   = 70 * 10 * 10    (note always g has value = 10 )

   = 7000 J

K.E = 1/2 mv2

   = 1/2 * 70 * 100      ($10^{2}$ = 100)

  = 1/2 * 7000

  = 3500 J    (7000 will be divided by 2 = 3500)

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