(10分)一个质量m=10kg的静止物体与水平地面间滑动摩擦系数μ=0.5,受到一个大小为100N与水平方向成θ=37°
1个回答

解题思路:

(1)由受力分析知

F

N

=

G

F

sin

37

=

40

N

(1

)

由摩擦力公式得

F

f

=

μ

F

N

=

20

N

(1

)

由牛顿第二定律

W

=

F

cos

37

F

f

=

m

a

(2

)

解得

a

=

6

m

/

s

2(1

)

由位移公式可得

x

=

=

3

m

(1

)

W

F

=

"

F

x

"

cos

37

=

240

J

(1

)

(2)

W

F

f

=

"

F

f

x

"

cos

180

=

60

J

(2

)

(3)

W

=

F

x

=

180

J

(1

)

(1)240J;(2)-60J;(3)180J

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