Question

A block of mass 5 kg slides down a rough inclined surface the angle of inclination is 45 degree the coefficient of sliding friction is 0.2 when the block slides 10m the work done on the block by force of friction is

Answer 1

here is ur answer mate..

please mark me brainlist

Answer 2

Answer:

$\sf{50\sqrt{2}\:Joules}$

Explanation:

Given:

• Mass (m) = 5 kg
• Angle = 45°
• $\sf{\mu}$ = 0.2
• Displacement (s) = 10 m

To find:

• Work done = ?

Solution:

We know, $\blue{\sf{Work\:done\:=\:Fscos\theta}}$ , friction is also a force so we also remember that

$\green{\sf{friction\:=\:\mu\:\times \:N}}$, Co-efficient of friction is given. We know, normal force (N) = mg.

The angle given is 45° so we know that $\sf{\cos45\:=\:\dfrac{\sqrt{2}}{2}}$

Let's find the friction first.

$\:$

$\Longrightarrow\:\sf{Friction\:=\:\mu\:\times\:mg}$

$\:$

$\Longrightarrow$ Friction = 0.2 × 5 × 10

$\:$

$\Longrightarrow$ Friction = 2 × 5

$\:$

$\therefore$ Friction = 10 N.

$\:$

Now, just put all the values in the formula of work done.

$\:$

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

$\:$

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

$\:$

$\small\Longrightarrow\underline{\boxed{\sf\purple{Work\:done\:=\:50\sqrt{2}\:J}}}$