Question

. On a rough horizontal surface, a body of mass 2 kg is
given a velocity of 10 m/s. If the coefficient of friction
is 0.2 and g = 10 m/s. the body will stop after
covering a distance of​

Answer 1

Answer:-

Case-1.

There is no external force acting only frictional force is acting.

$Normal \: reaction =mg$

$Normal reaction =2 \times 10 = 20 \: newton$

$friction = 0.2 \times 20 = 4 \: newton$

No external force this implies that

$friction = ma$

$4 = 2 \times m$

$a = 2 \frac{m}{ {s}^{2} }$

so, acceleration of the body is 2m/s^2.

Case-2

u = 10 m/s.

v = 0 m/s.

a = - 2 m/s^2.( retarded acceleration).

s = ?.

From third equation of motion

${v}^{2} - {u}^{2} = 2as$

substituting the values.

$- {10}^{2} = - 2 \times 2 \times s$

$s = \frac{100}{4}$

$s = 25 \: meters$

so, after travelling a distance of 25 meters the body will come to rest.

Answer 2

Work done by external force + Work done by conservative and non-conservative forces

= Change in Kinetic Energy

Here, external force is frictional force, which is = weight* coeff. Of friction

= 20*0.2 (assuming g=10 m/s2)

= 4 N

There are no other forces, so their work is obviously zero.

Change in KE= KE (final) - KE (initial)

= 0−12mv2

= −12mv2

Assuming body is moving towards +ve x direction, friction will be towards negative x direction.

Therefore,

work done by friction= 4d cos 180

because frictional force remains constant, both in magnitude as well as direction.

(d is the distance travelled by the body before it stops )

=> Work done by friction = -4 d.

=> −4d=−12mv2

=> 4d=12mv2

=>  4d=12(2)(20)2  

=> d=200m