Question

A block of Mass M is moving with a velocity v on a straight surface. What are the shortest distance and shortest time in which the block can be stopped if μ is a coefficient of friction
a.v2/2μg,v/μg
b. v2/μg,v/μg
c.v2/2Mg,v/μg
d none of the above

Answer 1

Explanation:

Force of friction opposes the motion

Force of friction=μN=μmg

Therefore retardation =μmg/m=μg

From v2=u2+2as

or

S=v2/2μg

from v=u+at

or t=v/μg

Answer 2

Concept:

• One-dimensional motion
• We will have to use kinematics equations
• Frictional forces

Given:

• Block of mass M
• The initial velocity of the block = v
• The final velocity of the block = 0 (the block is stopped)
• The coefficient of friction = μ

Find:

• the shortest distance in which the block is stopped
• the shortest time in which the block is stopped

Solution:

For the calculation of the shortest distance, we use the following kinematics equation,

v² = u² +2as

In this case, frictional force = μmg, acts in the opposite direction of the motion.

The acceleration due to the frictional force is = -μmg/m = -μg

0 = v² +2(-μg)s

2μgs = v²

s = v²/2μg

To calculate the shortest time, we use the following kinematics equation.

v = u+at

0 = v +(-μg)t

μgt = v

t = v/μg

The correct option is a). The shortest distance is v²/2μg and the shortest time in which the block comes to rest is v/μg.

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