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