Work is said to be done only when the force applied to the body produces a displacement in the body.
Mathematically,
Work done is by the force acting on a body is defined as the product of force and displacement of the body in the direction of force.
i.e. Work = Force x displacement
W = F x d.

To do work there must be motion of object with the application of force. But here, displacement of the building is zero.
i.e. W = F x 0 = 0
Hence, even if he gets tired, there is no work done.

1 J (1 Joule) work is said to be done if 1 Newton force displaces a body through a distance of 1 meter in the direction of force.

It is because it does not require any direction for its description.

Yes, it is possible that a force is acting on a body but still the work done is zero. For example, in case of a man pushing a hill.

There are two types of work done:

1. Work done against gravity: While lifting a body to a height, work is done against gravity.
2. Work done against friction: When a body is pushed or pulled then work is done against friction.

When force is applied perpendicular to the displacement, the angle θ = 90⁰.
Hence,
W = F s.cos90⁰ = 0
So, no work is done.

As we know that, W = F x d having unit Nm. So, in terms of fundamental unit it is kgm²/s².

1 erg work is the work done by a force of 1 dyne to move a body through a distance of 1 cm in the direction of force.

Work Done (W) = Force (F) x displacement (d)

Force (F) = 300N
Distance (d) = 70cm = 0.7m
Work done (W) = ?
By using formula, W = F x d
= 300 x 0. 7
= 210 J
Thus, a man does 210 J work.

Solution:
Here,
Mass (m) = 30kg
Height (h) =1.5km = 1500m
Acceleration due to gravity (g) = 10m/s²
As we know that,
W = mgh
or, W = 30 x 10 x 1500
or, W = 450000J
= 4.5 x 10⁵J

Solution:
Here,
Work done (W) = 50KJ = 50000J
Distance (d) = 10m
Force (F) = ?
By using formula,
W = F x d

or, 50000 = F x 10
or, F = 50000/10
or, F = 5000N

Here, Force (F) = 10N
Displacement (d) = 1m
Work done (W) = ?

We have,
W = F.d = 10 \(\times\) 1 = 10J
Therefore, work done by the force is 10J.

Here Mass (m₁) = 200 kg
Mass of the person (m₂) = 50 kg
Height (h) = 10 m
Acceleration due to gravity (g) = \(10m/s^{2}\)
We have,
W = F.d [ \(\therefore\) F = m.g]
= m.g.d [ \(\therefore\) m= m₁+m₂ = 200+50 = 250 kg]
= 250 \(\times\) 10 \(\times\) 10
= 2.5 \(\times\) \(10^{4}\)J

Thus, work done against the gravity is 2.5 \(\times\) \(10^{4}\)J

Given,
D = 20m
f = 200N
\(\theta\) = 15 \(^o\)

We have,
W = F.d (Cos \(\theta\) + Sin \(\theta\) )
= 200.20 (Cos 15 \(^0\) + Sin 15 \(^0\) )
= 200 \(\times\) 10 (0.9659 + 0.2588)
= 200 \(\times\) 10 \(\times\) 1.2247 = 2449.4J

Thus, work done on the slope is 2449.4J