Question

A stone weighing 1 kg and sliding on ice with a velocity of 2 m/s is stopped by friction in 10 s. The force of friction (assuming it to be constant) will be

Answer 1

Answer:

1/5 N

Explanation:

v= u + at

v = 0

u = 2m/s , t = 10s

Solve & get a = 1/5

Mass = 1 kg , so Ma = F = 1/5 N

Answer 2

The force of friction will be 0.2 N

Explanation:

It is given that,

Mass of he stone, m = 1 kg

Initial speed of the stone, u = 2 m/s

Finally it stops, v = 0

Time, t = 10 s

Let F is the force of friction. We know that the for is given by the product of mass and acceleration as ,

$F=ma$

$F=m\times \dfrac{v-u}{t}$

$F=-m\times \dfrac{u}{t}$

$F=-1\times \dfrac{2}{10}$

F = -0.018 N

or

F = -0.2 N

So, the force of friction will be 0.2 N. Hence, this is the required solution.

Learn more,

Second law of motion

https://brainly.in/question/4684417