一靜止於水平面之物體m=1kg

用戶37276

2013-01-10 8:56 am

一靜止於水平面之物體m=1kg 由靜止受到5N之水平作用力 2秒內之位移為6cm 若該物體改由靜止受定力8N作用則2秒內之位移為何

更新1:

1.s=vot+at平方/2 6=a/2*2的平方所以a=3 F-f=ma 5-f=1*3 f=2 2.F扑浪-f=ma扑浪 8-2=1*a扑浪 a扑浪=6 s=vot+1/2at平方 =1/2*6*2的平方 =12cm

回答 (1)

天同

2013-01-10 4:32 pm

✔ 最佳答案

Use equation of motion: s = ut + (1/2)at^2
with s = 6 cm = 0.06 m, u = 0 m/s, t = 2 s, a= ?
hence, 0.06 = (1/2)a(2^2)
a = 2x0.06/4 m/s^2 = 0.03 m/s^2

Use: net force = mass x acceleration
(5 - Ff) = 1 x 0.03 where Ff is the frictional force
Ff = (5 - 0.03) N = 4.97 N

When the object is acted by the 8 N force,
(8 - 4.97) = 1 x a' where a' is the acceleration of the object
a' = 3.03 m/s^2

Apply equation of motion: s = ut + (1/2)at^2
with u = 0 m/s, t = 2 s, a = 3.03 m/s^2, s =?
hence, s = (1/2).(3.03).(2^2) m = 6.06 m

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