物理 : 關於動能的問題

(a) I suppose the unit for velocity on the graph is in "m/s".
Hence, kinetic energy (動能) on reaching the floor
= (1/2).(0.2).(5^2) J = 2.5 J

(b) Assume no energy loss during rebound, by Conservation of Energy,
Gain in potential energy = loss of kinetic energy
i.e. 0.2g.h = 2.5
where h is the maximum height reached by the ball, and g is the acceleration due to gravity, taken to be 10 m/s^2
h = 1.25 m

(c) First assume there is energy conservation (能量守恆), then
(1/2).(0.2)v^2 + (0.2)gh = K
where v is the velocity of the ball at height h, K is a constant
h = K/0.2g - v^2/2g
or h = ho - v^2/2g where ho = K/0.2g, is the height where the ball is released

But v = gt
hence, h = ho - (gt)^2/2g
i.e. h = ho - 5t^2 --------------- (1)

Therefore, if the given displacement-time graph (位移-時間曲線) follows equation (1), energy conservation is verified.

快去這里*****我每次都是去這里看
主卍

(a) max(KE) = (1/2) mv² = (1/2) 0.2 * 5² = 2.5J

(b) max(PE) = mgh = max(KE)
0.2 * 10 * h = 2.5
h = 1.25