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Chapter 6: Q.54 (page 157)

A large box of mass Mis moving on a horizontal surface at speed V0 . A small box of mass m sits on top of the large box. The coefficients of static and kinetic friction between the two boxes are μsand μ1, respectively. Find an expression for the shortest distancedmin in which the large box can stop without the small box slipping.

Short Answer

Expert verified

The expression for the minimum distance dminfor which the large box can stop without the slipping of a small boxdmin=v022(9.8)μμ.

Step by step solution

01

Given.

Speed of the large box: v0

Coefficient of static friction: μs

Coefficient of static friction: μk

02

Formula used.

F=μnv2=u2+2as

03

Calculation.

There are 3 forces acting on the small box

  1. Normal force, n=mg
  2. Force due to gravity.

Friction force Ff

We have to find the forces in horizontal direction, so forces in x- direction would be

Ff=ma=mgμs⋯. (1)

According to the equation of kinematics, v2=v02+2ad

The final velocity of the large box is zero because the box stops. So the equation become v02=2ad.

Now solving for dby substituting the value of 'a' from the equation (1).

We get

v02=2add=v022(9.8)μx

The minimum distance is dmin=v022(9.8)μx.

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