A stone of mass 64.0 g is thrown vertically upward from the ground with an initial speed of 20.0 ms-1. The gravitational potential energy at the ground level is considered to be zero. Applying the principle of conservation of energy, the potential energy attained the total energy of the stone at its halfway point can be calculated to be: (g= 10 m s-2).
A stone of mass 64.0 g is thrown vertically upward from the ground with an initial speed of 20.0 ms-1. The gravitational potential energy at the ground level is considered to be zero. Applying the principle of conservation of energy, the potential energy attained the total energy of the stone at its halfway point can be calculated to be: (g= 10 m s-2).
The correct option is A 12.8 J
Given : m = 64.0 g = 64/1000 kg
v = 20.0 ms-1 and g = 10 ms-2
Kinetic energy at the bottom point or the point of throwing is:
k = 1/2 mv2 = 1/2 x 64/1000 x (20.0)2 = 12.8 J
By the principle of conservation of energy, the kinetic energy at the ground level is totally converted into potential energy at the maximum height. Potential energy at the maximum height, U= 1.28 x 104J.
The total energy of the body will remain conserved throughout the motion and hence Total energy anywhere is equal to PE at the maximum height or KE at the lowest position. Thus TE at halfway point is 12.8J
12.8 J
Given : m = 64.0 g = 64/1000 kg
v = 20.0 ms-1 and g = 10 ms-2
Kinetic energy at the bottom point or the point of throwing is:
k = 1/2 mv2 = 1/2 x 64/1000 x (20.0)2 = 12.8 J
By the principle of conservation of energy, the kinetic energy at the ground level is totally converted into potential energy at the maximum height. Potential energy at the maximum height, U= 1.28 x 104J.
The total energy of the body will remain conserved throughout the motion and hence Total energy anywhere is equal to PE at the maximum height or KE at the lowest position. Thus TE at halfway point is 12.8J
