Question

A $2kg$ mass starts from rest on an inclined smooth surface with inclination ${30}^o$ and length $2m$. How much will it travel before coming into rest on a horizontal frictional surface with frictional coefficient of $0.25$

• A. 4 M
• B. 6 M
• C. 8 M
• D. 2 M

Answer 1

The correct option is A 4 M

Given that,

Length of inclined plane$=2m$

We know that,

The initial velocity, $u=0$

The final velocity, $v=v$

Then,

$v^2=u^2+2as$

$v^2=u^2+2g\sin \theta \times 2$

$v^2=0+2\times 10\times \frac{1}{2}\times 2$

$v^2=20m/s$

Now, the frictional force is

$F_F=\mu N$

$ma=\mu mg$

Now, the acceleration is

$a=\frac{\mu mg}{m}$

$a=\frac{0.25\times 2\times 10}{2}$

$a=2.5m/s^2$

Now, the distance traveled before coming to rest

From equation of motion,

$v^2=u^2+2as$

$0=20-2\times 2.5\times s$

$s=4m$

Hence, it will travel $4$m before coming to rest.