Question

A 25.0-kg box of textbooks rests on a loading ramp that makes an angle α with the horizontal. The coefficient of kinetic friction is 0.25, and the coefficient of static friction is 0.35. (a) As α is increased, find the minimum angle at which the box starts to slip. (b) At this angle, find the acceleration once the box has begun to move. (c) At this angle, how fast will the box be moving after it has slid 5.0 m along the loading ramp?

Answer 1

Answer:

Explanation:

a ) When the box starts to slip

static friction = mg sinα

mg cosα x μ =  mg sinα (  μ is coefficient of static friction )

Tanα =  μ  = .35

α  = 19.2°

b ) Once box starts moving , kinetic friction will start applying on it

kinetic friction = mg cos19.2 x .25

= 2.31m

net force downward = mgsin19.2 - mgcos19.2 x .25

= m ( 3.22 - 2.31 )

= .91 m

acceleration downward ( a )   = .91 m / s²

c )

v² = u² + 2 a s

= 0 + 2 x .91 x 5

= 9.1

v = 3 m / s

Answer 2

(a) The angle will be "19.2°".

(b) The acceleration will be "0.91 m/s²".

(c) The speed will be "3 m/s".

According to the question,

• Mass = 25.0 kg
• Kinetic friction = 0.25
• Coefficient of static friction = 0.35

(a)

We know,

→ $Static \ friction = mg \ Sin \alpha$

→ $mg Cos \alpha\times \mu = mg Sin \alpha$

then,

→ $Tan \alpha = \mu = 0.35$

                $\alpha = 19.2^{\circ}$

(b)

Kinetic friction = $mg Cos 19.2^{\circ}\times 0.25$

                         = $2.31 \ m$

and,

Net force downwards = $mgSin 19.2^{\circ} - mg Cos 19.2\times 0.25$

By substituting the values, we get

                                  = $m (3.22-2.31)$

                                  = $0.91 \ m/s^2$

(c)

The speed will be:

→ $v^2 = u^2 +2as$

        $= 0+2\times 0.91\times 5$

        $= 9.1$

     $v = 3 m/s$

Thus the responses above answer correct.                                

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