Question

The diagram shows a barrel of weight $1.0 \times 10^{3} \,N$ on a frictionless slope inclined at $30°$ to the horizontal. The force -is parallel to the slope. What is the work done in moving the barrel a distance of $5.0 \,m$ up the slope?

• A. $2.5 \times 10^{3} \,J$
• B. $4.3 \times 10^{3} \,J$
• C. $5.0 \times 10^{3} \,J$
• D. $1.0 \times 10^{3} \,J$

Answer 1

Answer: (A)

Work done in moving the barrel on the frictionless slope= change in potential energy of barrel
$W = mg (h_2 - h_1)$
Here, $mg = 1.0 \times 10^3\, N$ $(h_2 - h_1) = s \, \sin \,30° = 5 \, sin\, 30°$ = 2.5 m
$\therefore \:\:\: W = 1.0 \times 10^3 \times 2.5 = 2.5 \times 10^3\, J$