Laws Of Motion Question 10
Question 10 - 2024 (30 Jan Shift 2)
A block of mass $1 kg$ is pushed up a surface inclined to horizontal at an angle of $60^{\circ}$ by a force of $10 N$ parallel to the inclined surface as shown in figure. When the block is pushed up by $10 m$ along inclined surface, the work done against frictional force is : $\left[g=10 m / s^{2}\right]$
(1) $5 \sqrt{3} J$
(2) $5 J$
(3) $5 \times 10^{3} J$
(4) $10 J$
Show Answer
Answer: (2)
Solution:
Work done again frictional force
$=\mu N \times 10$
$=0.1 \times 5 \times 10=5 J$
