A block of mass 1 kg slides down a curved track that is one quadrant of a circle of radius 1 m. Speed of the block at the bottom is 2 m / s. Work done by the frictional force on the block when it reaches at the bottom is

Question

A block of mass 1 kg slides down a curved track that is one quadrant of a circle of radius 1 m. Speed of the block at the bottom is 2 m / s. Work done by the frictional force on the block when it reaches at the bottom is (A) 8 J (B) 4 J (C) -8 J (D) 0 J

Tardigrade Admin · Accepted Answer

Given, m =1 kg, v =2 m / s R =1 m W =? From energy conservation, m g h+ W=(1/2) m v2 =(1/2) m v2-m g h =(1/2) × 1(2)2-1 × 10=-8 J

Q. A block of mass $1 k g$ slides down a curved track that is one quadrant of a circle of radius $1 m$ . Speed of the block at the bottom is $2 m / s$ . Work done by the frictional force on the block when it reaches at the bottom is

8 J

4 J

-8 J

0 J

Given, $m = 1 k g$ ,
$v = 2 m / s$
$R = 1 m$
$W = ?$
From energy conservation,
$m g h + W = \frac{1}{2} m v^{2}$
$= \frac{1}{2} m v^{2} - m g h$
$= \frac{1}{2} \times 1 (2)^{2} - 1 \times 10 = - 8 J$

Q. A block of mass 1kg slides down a curved track that is one quadrant of a circle of radius 1m. Speed of the block at the bottom is 2m/s. Work done by the frictional force on the block when it reaches at the bottom is

8 J

4 J

-8 J

0 J

Given, m=1kg, v=2m/s R=1m W=? From energy conservation, mgh+W=21​mv2 =21​mv2−mgh =21​×1(2)2−1×10=−8J

Q. A block of mass $1 k g$ slides down a curved track that is one quadrant of a circle of radius $1 m$ . Speed of the block at the bottom is $2 m / s$ . Work done by the frictional force on the block when it reaches at the bottom is

Given, $m = 1 k g$ ,
$v = 2 m / s$
$R = 1 m$
$W = ?$
From energy conservation,
$m g h + W = \frac{1}{2} m v^{2}$
$= \frac{1}{2} m v^{2} - m g h$
$= \frac{1}{2} \times 1 (2)^{2} - 1 \times 10 = - 8 J$