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

Answer 1

Answer: (A)

$\ce{2.5 \times 10^{3} \,J}$

Answer 2

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
$\therefore \:\:\: W = 1.0 \times 10^3 \times 2.5 = 2.5 \times 10^3\, J$