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The diagram shows a barrel of weight $1.0 \times 10^{3} \mathrm{~N}$ on a frictionless slope inclined at $30^{\circ}$ to the horizontal.
The force is parallel to the slope. What is the work done in moving the barrel a distance of $5.0 \mathrm{~m}$ up the slope?
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The force is parallel to the slope. What is the work done in moving the barrel a distance of $5.0 \mathrm{~m}$ up the slope?
Solution:
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Verified Answer
The correct answer is:
$2.5 \times 10^{3} \mathrm{~J}$
The given situation is shown below
Work done in moving the barrel on the frictionless slope is equal to change in potential energy.
i.e. $W=m g\left(h_{1}-h_{2}\right)$
Here, $\quad m g=1 \times 10^{3} \mathrm{~N}$
From figure, $\sin 30^{\circ}=\frac{h_{1}-h_{2}}{5}$ $\Rightarrow \quad h_{1}-h_{2}=5 \sin 30^{\circ}=\frac{5}{2}=25 \mathrm{~m}$
Putting these values in Eq. (i), we get
$$
W=1 \times 10^{3}(2.5)=2.5 \times 10^{3} \mathrm{~J}
$$
Work done in moving the barrel on the frictionless slope is equal to change in potential energy.
i.e. $W=m g\left(h_{1}-h_{2}\right)$
Here, $\quad m g=1 \times 10^{3} \mathrm{~N}$
From figure, $\sin 30^{\circ}=\frac{h_{1}-h_{2}}{5}$ $\Rightarrow \quad h_{1}-h_{2}=5 \sin 30^{\circ}=\frac{5}{2}=25 \mathrm{~m}$
Putting these values in Eq. (i), we get
$$
W=1 \times 10^{3}(2.5)=2.5 \times 10^{3} \mathrm{~J}
$$
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