Question

A box of mass 4 kg slides down a slope. The initial speed of the box was 4 m/s. After covering a distance of 20cm it comes to rest due to the friction between box and surface. Find work done by frictional force.​

Answer 1

Answer:

• 32 Joules is the required work done by the frictional force.

Explanation:

According to the Question

It is given that,

• Mass ,m = 4kg
• Initial velocity ,u = 4m/s
• Final velocity ,v = 0m/s
• Distance covered , s = 20cm

we have to calculate the work done by the frictional force.

Firstly we calculate the acceleration of the box .

Using Kinematic Equation

• v² = u² - 2as

↠ 0² = 4² - 2 × a × 0.2

↠ 0 = 16 - 0.4 × a

↠ -16 = -0.4 × a

↠ a = 16/0.4

↠ a = 40m/s

Now, calculating the work done by the frictional force .

• Work Done = Force × Displacement

On substituting the value we get

↠ Work Done = 4 × 40 × 0.2

↠ Work Done = 160 × 0.2

↠ Work Done = 32J

• Hence the work done by the frictional force is 32 J .

Answer 2

Given:-

• Mass, m= 4 kg
• Initial velocity, u = 4m/s
• Distance, s = 20 cm = ( in m ) 0.2 m
• Since it is given that after covering 20 cm distance, box comes to rest so , final velocity, v = 0 m/s

To Find:-

• Work done by frictional force

First we calculate the acceleration by using third law of motion

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \sf \large{\boxed{ \red{ \sf{v² – u² = 2as}}}}$

Now Substituting the values,

$\: \: \: \implies \sf{ \bold{0² – 4² = 2×a×</strong><strong>0</strong><strong>.</strong><strong>2</strong><strong>}}$

$\: \: \: \: \: \: \: \: \: \: \implies \sf{ \bold{0 - 16 = 0.4a}}$

$\: \: \: \: \: \: \: \: \: \: \implies \sf{ \bold{a = \frac{ - 16}{0.4}}}$

$\: \: \: \: \: \: \: \: \: \: \implies \sf{ \bold{a = \frac{ - 16 \times 10}{4} }}$

$\: \: \: \: \: \: \: \: \: \: \implies \sf{ \bold{a = - 40}}$

Negative ( – ) sign shows retardation.

Now find the work done,

$\: \: \: \: \: \: \: \: \: \: \sf{ \bold{ \sf Work, W = FS}}$

Now put the values,

$\: \: \: \: \: \: \: \: \: \: \: \implies \sf \bold{W = ( m × a ) × s}$

$\: \: \: \: \: \: \: \: \: \: \: \implies \sf \bold{W = 4 \times ( - 40) \times 0.2}$

$\: \: \: \: \: \: \: \: \: \: \: \implies \sf \bold{W = - 16 \cancel0 \times \frac{02}{ \cancel10} }$

$\: \: \: \: \: \: \: \: \: \: \: \implies \sf \bold{W = - 32 \: Nm}$

Here, Negative (–) sign shows negative work

Additional information:-

•Work done is said to be negative if the applied force is in opposite direction to that of displacement. In this case angle between force and displacement is 180°.

Examples:-

• Work done by frictional force of air by rough surface when an object is lifted from the surface of the Earth.
• Work done by friction on an object sliding/ slope on a rough surface.