Work done by the force acting on a body is defined as the product of force and displacement of the body in the direction of the force. It is a scalar quantity. Force acting on the body must produce a displacement for the work is to be done by the force. Thus for the work to be done, the following two conditions must be fulfilled:
Mathematically,
Work= Force × Displacement (in the direction of the force)
i.e. W= F.d ................(i)
The SI unit of force is newton (N) and that of displacement is a metre (m). So, the unit of work is Newton metre (Nm), which is Joule (J).
Thus, one joule of work is said to be done when one Newton force displaces a body through one metre in its own direction.
Work against friction
Force is to be applied to move, roll, or drag a body over a surface of another body. Example: If an object is dragged to a certain distance (d).
Activity: To demonstrate work against friction.
At first, placed the brick over the surface of the table. Now, pulled the brick from initial position to final position as shown in the figure.It moves by a displacement's'. Here, force is applied against the frictional force between the surface of the brick and the table to set the brick in motion. Here, the work is done against friction.
Work against gravity
Force is applied to lift a body against the force of gravity. Example: If an object is lifted to a certain height (h).
Activity: To demonstrate the work against gravity.
Take a body of mass 'm' and a spring balance to lift the body. Then lift the body of mass 'm' upto the certain height 'h'. The pointer of the spring balance gives the force applied 'F' against the gravity. Here, the work is done against the gravity and it is given by the product of the force 'F' and height 'H'.
On the basis of the direction of motion of a body and the direction of force applied, the way of calculation of work done is determined.
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:
When force is applied perpendicular to the displacement, the angle θ = 90^{0}.
Hence,
W = F s.cos90^{0} = 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 kgm^{2}/s^{2}.
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/s^{2}
As we know that,
W = mgh
or, W = 30 x 10 x 1500
or, W = 450000J
= 4.5 x 10^{5}J
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 (m_{1}) = 200 kg
Mass of the person (m_{2}) = 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= m_{1}+m_{2} = 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
SI Unit of work is ------.
watt
mole
horse
joule
if a man lifts 200 kg of water from a well 10m deep what is the work done by the man ?
_{}10^{4 }J
10^{5} J
10^{6 }J
2 X 10^{4 }J
What is the formula related to velocity?
p=w × v
p=v/w
p=f × v
p=v/t
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why work is called scar quantity?
Why work is called scar quantity?
Mar 21, 2017
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