A man drags a block through 10 m on rough surface (μ = 0.5). A force of √3 kN acting at 30° to the horizontal. The work done by applied force is

Question

A man drags a block through 10 m on rough surface (μ = 0.5). A force of √3 kN acting at 30° to the horizontal. The work done by applied force is (A) zero (B) 7.5 kJ (C) 15 kJ (D) 10 kJ

Tardigrade Admin · Accepted Answer

Horizontal component of applied force =(√3 × 103) × cos 30° =√3 × 103 × (√3/2) =(3/2) × 103 N Work done = F.s =(3/2) × 103 × 10 =15 × 103 J =15 kJ

Q. A man drags a block through $10 m$ on rough surface $(μ = 0.5)$ . A force of $3$ kN acting at $3 0^{\circ}$ to the horizontal. The work done by applied force is

zero

7.5 kJ

15 kJ

10 kJ

Horizontal component of applied force
$= (3 \times 1 0^{3}) \times cos 3 0^{\circ}$
$= 3 \times 1 0^{3} \times \frac{3}{2}$
$= \frac{3}{2} \times 1 0^{3} N$
Work done $= F . s$
$= \frac{3}{2} \times 1 0^{3} \times 10$
$= 15 \times 1 0^{3} J$
$= 15 k J$

Q. A man drags a block through 10m on rough surface (μ=0.5). A force of 3​ kN acting at 30∘ to the horizontal. The work done by applied force is