Question

A force of 25N is acting on an object which is inclined at 45° . The object is displaced through 20 metre in the direction of force . What is the work done in this case?
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Answer 1

Given:-

• Force ,F = 25N

• Displacement ,s = 20m

• Cosθ = 45°

To Find:-

• Work done ,W

Solution:-

As we know that work done is defined as the product of force and displacement. Here Work Done

• W = Fs × Cos45°

where,

F is the Force

s is the displacement

Cos45° = 1/√2

Substitute the value we get

→W = 25×20 × Cos45°

→ W = 500 × 1/√2

→ W = 353 .55 J

Therefore,the Work Done in this case 353 .55 Joules.

Answer 2

Answer:

Given :-

• A force of 25 N is acting on an object which is inclined at 45° .
• The object is displayed through 20 m in the direction of force.

To Find :-

• What is the work done.

Formula Used :-

$\boxed{\bold{\small{Work\: Done\: =\: Force\: ×\: Displacement\: ×\: cos45°}}}$

Solution :-

Given :

• Force (F) = 25 N
• Displacement (s) = 20 m
• cos45° = $\dfrac{1}{\sqrt{2}}$

According to the question by using the formula we get,

⇒ Work Done = 25 × 20 × $\dfrac{1}{\sqrt{2}}$

⇒ Work Done = 500 × $\dfrac{1}{\sqrt{2}}$

➠ Work Done = 353.55 J

$\therefore$ Work done by the object is 353.55 J .