A block of mass 10 kg slides down a rough slope which is inclined at 45° to the horizontal. The coefficient of sliding friction is 0.30. When the block has slide 5 m, the work done on the block by the force of friction is nearly

Question

A block of mass 10 kg slides down a rough slope which is inclined at 45° to the horizontal. The coefficient of sliding friction is 0.30. When the block has slide 5 m, the work done on the block by the force of friction is nearly (A) 115 J (B) 75 √2 J (C) 321.4 J (D) -321.4 J

Tardigrade Admin · Accepted Answer

As, F=μ m g cos θ orF=0.30 × 10 × 10 cos 45° orF=(30/√2) N W=F × s=(30/√2) × 5 =(150/√2) × (√2/√2)=75 √2 J This is negative work because F and s are oppositely directed.

Q. A block of mass $10 k g$ slides down a rough slope which is inclined at $4 5^{\circ}$ to the horizontal. The coefficient of sliding friction is $0.30$ . When the block has slide $5 m$ , the work done on the block by the force of friction is nearly

115 J

$752 J$

321.4 J

-321.4 J

As, $F = μ m g cos θ$
or $F = 0.30 \times 10 \times 10 cos 4 5^{\circ}$
or $F = \frac{30}{2} N$
$W = F \times s = \frac{30}{2} \times 5$
$= \frac{150}{2} \times \frac{2}{2} = 752 J$
This is negative work because $F$ and $s$ are oppositely directed.

Q. A block of mass 10kg slides down a rough slope which is inclined at 45∘ to the horizontal. The coefficient of sliding friction is 0.30. When the block has slide 5m, the work done on the block by the force of friction is nearly