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# A physics Question (Urgent)

Show the calculation steps very clearly. And explanation is necessary. (Use diagram to cooperate with the explanation words if you can.)

1. A 4kg wooden block rests on a friction-compensated inclined plane with inclined angle Ө. Suppose the friction between the block and the inclined plane is 10N.

(a) Find the inclined angle Ө.

(b) If the block is given a sudden downward jerk so that it moves with 5m/s, how much will the block displace in 1.2s?

(c) If the block is given a sudden upward jerk so that it moves with 5 m/s, how much will the block displace in 1.2s?

how about Q. 1c)???

1c) Ans: 2.4m

(5)(1.2) (0.5)(-5)(1.44) is not equal to 2.4m.

Some questions about 1c).

Why the force is in negative sign???? and what is the final velocity????

### 1 Answer

- 1 decade agoFavorite Answer
............................. /

(wooden block) 口/

........................... / Ө

.......................... /_____________

1.(a) The weight of the wooden block is 4X10=40N

The compoents of the weight of block downward = friction (since it rests on the plane)

sinӨ x 40 = 10

Ө = 14.5'

(b) Since the compoents of the weight of block downward = friction

by newton's first law,

every object remains at uniform motion or state of reast unless acted by a net force

so that the block will move at constant velocity

s=1/2x(u+v)xt

s=1/2(5+5)X1.2

s=6m

Hope that it can help =]

2007-12-15 22:34:17 補充：

(c)There is a force acting the block upwardit will be canceled with compoent of weightThe force acting upward :F = maF = 4 x 5F = 20N

2007-12-15 22:34:33 補充：

The net force acting on the block without friction:20N - 10N = 10N (upward)So the direction of friction will be downward (since the friction must be opposite to the motion)

2007-12-15 22:34:53 補充：

The net force of the block with friction:10N - 10N = 0NSo that there is no net force acting on the blockThe block will remain its state of rest.

2007-12-16 02:15:12 補充：

Sorry that i have made a mistakeThe friction is 10N(downward) and the compoent of weight is 10N(downward)The net force acting to the block:10N 10N = 20N(downward)

2007-12-16 02:15:23 補充：

The deceleration of the block:Take upward to be possitiveF = m x a-20 = 4 x a a = -5m/s^2s = ut (1/2) x at^2s = (5)(1.2) (0.5)(-5)(1.44)s = 2.4mHope they can help !

2007-12-16 12:18:11 補充：

There is a '＋' side between themthe yahoo system has such problem that disappear my '＋' sidethe equation should be (5)(1.2)＋(0.5)(-5)(1.44)

2007-12-16 17:46:46 補充：

Why the force is in negative sign???? Since the movement of the block is 5m/s upward,so the friction is downward 10N,and the compoent of weight won't be canceled with the friction,

2007-12-16 17:46:55 補充：

the compoent of weight will also be downward 10N,finally there is a downward net force 20N acting on the block.

2007-12-16 17:48:43 補充：

what is the final velocity???? you may use the answer in (c) to calculate,but what is the meaning of final?is it 1.2s or after 1.2s ?but the final velcity must be zero finally.

Source(s): ME