Question

A small block slides along a path, without friction, until the block reaches the section $L=3m$, which begins at height $h=3m$ on a flat incline of angle ${37}^o$, as shown. In that section, the coefficient of kinetic friction is ${\mu }_k$=$0.50$. The block passes through point $A$ with a speed of $\sqrt{136}m/s$. Find the speed (in m/s) of the block as it passes through the point $B$ where the friction ends. (Take $g=10m/s^2$)

• A. $4m/s$
• B. $6m/s$
• C. $8m/s$
• D. $10m/s$

Answer 1

The correct option is B $6m/s$

According to the conservation of total energy

Mechanical energy = kinetic energy ₊ potential energy=Constant

Let, m be the mass of a block

PE at point A

$=mgh$

$=m\times 10\times 0=0$

KE at point A=Total mechanical energy at point A

$=\frac{1}{2}m{v}^2$

$=\frac{1}{2}m{\left(\sqrt{136}\right)}^2$

$=68m$

Let at height 3 m on inclined plane, a point C is there.

Potential energy at point C,

$P{E}_C=m\times 10\times 3=30m$

Net Kinetic energy at point C

$=68m-30m$

$=38m$

While the travels from C to b, There will some loss of mechanical energy due to friction which iwill given by :

$E=\mu N\times 3$

Normal force

$N=mg\cos \theta$

$=m\times 10\times \cos \left(37\right)$

$=8.0m$

Kinetic friction force

$=\mu N$

$=0.5\times 8m$

$=4m$

Loss in total energy adue to friction

$4\times 3=12J$

Loss of kinetic energy due to the height

$=mgL\sin \theta$

$=m\times 10\times \sin \left(37\right)$

$=6mJ$

Net Kinetic energy at B

$=38m-12m-6m$

$=20mJ$

$so,\frac{1}{2}m{v}^2=20m$

$v=6.3m/s$

Therefore, the velocity of the body at point B will be approximate $6m/s$.