Question

A block of mass 5 KG slides down a rough inclined surface the angle of inclination is 45o the coefficient o
f sliding friction is 0.2 when the block
slides 10m the
workd one on the block by force of friction is.​

Answer 1

Answer:

• $50\sqrt{2}J$

Explanation:

$Given:-$

• Mass = 5 kg

• Angle of inclination = 45°

• $\sf{\mu}$ = 0.2

• Displacement = 10 m

$ToFind:-$

• Work done

$Solution :-$

$Friction = \mu \times mg$

$\implies \:\: Friction = 0.2 \times 5 \times 10$

$\implies \:\: Friction = 2 \times 5$

$\implies \:\: Friction = 10\:N$

$Work \ done = Fscos\:\theta$

$\implies \:\:Work \ done = 10 \times 10 \times cos45^{o}$

$\implies \:\:Work \ done =10 \times 10 \times \dfrac{\sqrt{2}}{2}$

$\implies \:\:Work \ done = \dfrac{100\times \sqrt{2}}{2}$

$\implies \:\:Work \ done = 50\sqrt{2}J$

Hence, The work done on the block by force of friction is $50\sqrt{2}J$