Subject: Science
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.
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.
A porter carrying a load of 30kg climbs up a mountain having vertical height of 1.5km from the ground. Calculate the work done against gravity (g = 10m/s2).
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
How much work is done by a force of 10N while moving an object through a distance of 1m in the direction of the force?
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.
Calculate the amount of work done by a person while taking a bag whose mass is 200kg to the top of a building of height 10m. (Use g= \(10m/s^{2}\). The mass of the person is 50kg.
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
A trolley is pulled on a slanted surface by applying an effort os 200N. If the length of the slope is 20m and the angle between the horizontal surface and inclined slope is 15 \(^o\) calculate the total work done.
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
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