[解决]Physic 问题，各位请帮忙解答。

jennykuwn

A crate of mass 50kg slides down a 30' incline . The crate's acceleration is 2.0ms-2, and the incline is 10m long..

a) what is the kinetic energy of the crate as it reaches the bottom of the incline? ANS：1000J

b) How much work is spent in overcoming friction?   ANS:  1450J
我想要看步骤。。谢谢

jennykuwn

老师给的答案是1000J，我算不到。拜托了。＞＜

数学系

To #1

Take g = 9.81 ms^-2. We have to consider the component of acceleration parallel to the surface now. Since the incline is 30', a_horizontal = g cos 60'.

Then, we have to consider the resultant acceleration a_r and final velocity v:
a_r = 2 + g cos 60' = 6.905 ms^-2

Apply the formula v^2 = u^2 + 2sa_r = 0 + 2(10)(6.905) = 138.1m^2s^-2

Thus, the kinetic energy is 0.5mv^2 = 0.5(50)(138.1) = 3452.5 J

jennykuwn

不过老师给的答案是1000J。

数学系

老师不一定队，我也不一定错。

数学系

我知道问题在那里了。

Let a = 2ms^-2. Then apply the formula v^2 = u^2 +2as = 0 + 2(2)(10) = 40 m^2s^-2

To find the kinetic energy, use the formula 0.5mv^2 = 0.5(50)(40) = 1000 J

数学系

我以为 a = 2 ms^-2 那辆破车的加速度，所以我还加上 g cos60'，以计算出那辆车的resultant acceleration。我误解题目意思，对不起。

For b) To find the friction of the surface f, use the formula ma = mg cos60' - f, thus we have f = mg cos60' - ma = 145 N (take g = 9.8 ms^-2). Thus, the work spent on overcoming friction is W = fs = 145(10) = 1450 J.

Lov瑜瑜4ever

mg sin 30 - friction force,f =ma
50x9.81x0.5 - f =50x2
                      f =245.25-100
                      f =145.25

work done=fs
               =145.25x10
               =1450J

jennykuwn

我以为 a = 2 ms^-2 那辆破车的加速度，所以我还加上 g cos60'，以计算出那辆车的resultant acceleration。 ...
数学系 发表于 1-8-2010 12:11 AM

噢，是我没有写好问题。。你们肯写下步骤让我参考，我已经很感激了。谢谢