The diagram shows a barrel of weight 1.0 × 103 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?

Question

The diagram shows a barrel of weight 1.0 × 103 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 × 103 J (B) 4.3 × 103 J (C) 5.0 × 103 J (D) 1.0 × 103 J

Tardigrade Admin · Accepted Answer

Work done in moving the barrel on the frictionless slope= change in potential energy of barrel

W = m g (h_{2} - h_{1})

Here,

m g = 1.0 \times 1 0^{3} N

(h_{2} - h_{1}) = s sin 30° = 5 s in 30°

= 2.5 m

∴ W = 1.0 \times 1 0^{3} \times 2.5 = 2.5 \times 1 0^{3} J

Q. The diagram shows a barrel of weight $1.0 \times 1 0^{3} N$ 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?

$2.5 \times 1 0^{3} J$

$4.3 \times 1 0^{3} J$

$5.0 \times 1 0^{3} J$

$1.0 \times 1 0^{3} J$

Work done in moving the barrel on the frictionless slope= change in potential energy of barrel
$W = m g (h_{2} - h_{1})$
Here, $m g = 1.0 \times 1 0^{3} N$ $(h_{2} - h_{1}) = s sin 30° = 5 s in 30°$ = 2.5 m
$∴ W = 1.0 \times 1 0^{3} \times 2.5 = 2.5 \times 1 0^{3} J$

Q. The diagram shows a barrel of weight 1.0×103N 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 5.0m up the slope?

2.5×103J

4.3×103J

5.0×103J

1.0×103J