Question

At the top of a frictionless inclined plane, a 0.50-kilogram block of ice possesses 5.0 joules of gravitational poteRtial energy with respect to the bottom of the incline. After sliding halfway down the plane, the block's gravitational potential energy is _______.

Answer 1

Answer:

after sliding halfway down the plane ( at 0.1275 m from the bottom) the potencial gravitational energy is 2.5 J

Explanation:

the gravitational potential energy with respect to the bottom is

E = m * g * h

h = E/ (m*g)

replacing values

h= 5 J / (0.5 kg * 9.8 m/s²) = 0.255 m

also

h = d * sin Ф , where d= inclined plane distance , Ф = angle with respect to the horizontal plane

for halfway down the plane d2 =d/2 ,

since h2=d2* sin Ф = (d/2) * sin Ф = (d* sin Ф)/2 = h/2

h2=h/2 = 0.255 m/2 = 0.1275 m

thus

E2 = m*g*h2=  (m*g*h)/2 = E/2 = 5 J /2 = 2.5 J

the potential energy E2 is 2.5 J