An object of certain mass is dropped from the top of a building of height 10m. At a certain point on its path, the ratio of its potential energy is found to be 1:3. The work done by the gravitational force of the earth in moving the body up to this point is 225 J. Determine the mass and the kinetic energy when it reaches the ground.[Take g=10m/s]
(Please give proper explanation)
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Answer:
Given –
- Height = 10 m
- At a certain point in its path, P.E. = 1:3
- The Work done by gravitational force is 225 J
- g = 10 m/s²
To Find –
- Mass of the object
- Kinetic energy when it reaches ground
Solution –
Potential energy = PE1 = mgh.
Potential energy at a given point will be : PE2 = ¼ * mgh.
Work Done is said to be the difference of the potential energy.
W = PE1 - PE2
W = mgh - ¼*mgh
W = 3/4 * mgh
Hence, Mass of the object :-
Mass = W ÷ (3/4*g*h)
Mass = W ÷ (0.75*g*h)
Mass = 225 ÷ (0.75*10*10)
Mass = 225 ÷ 75
Mass = 3 kg
Form third Equation of motion : v² - u² = 2gs.
As, ut is a falling object, u = 0.
Hence, v² = 2gs.
So, v² = 2*10*10
v² = 200
v = 14.142 m/s
Kinetic Energy : ½*mv²
KE = ½*3*(14.142)²
KE = ½*3*200
KE = 3*100
KE = 300 J
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