Work Done Quiz with answers

(a)

(b)

(a)

Work done, $W= F\times d = 50\times 4=200Joules$
It is important to note that when we are calculating work, the mass of the object that is being displaced is irrelevant.

(a)

From the figure, we can see that force is acting at an angle of 60 degrees with the direction of the displacement. Here force would have two components

1. x-axis
2. y-axis

The component of force along the y-axis does not contribute to the work done by the force. Only a component of force along the direction of displacement is effective. So we have,
$W=(F\cos\theta)d$
The force is applied at a $60$ angle from the direction of displacement, so $\theta = 60^0$
$W=(F\cos\theta)d$
$W=(10N)(\cos60^0)(20m)=(10N)(0.5)(20m)$
$W=100J$

(c)

It is given in the question that
$m=15 Kg$; $distance\;\;s= 2.0m$ ; $g=9.8 m/s^2$ and we have to calculate work $W=?$
Now, Force applied $=F=mg$ (weight of the box)
$F=15Kg\times9.8m/s^2=147N$
Work done $W=F\cdot s=147N\cdot 2.0 m = 294N$

(d)

We know that work done,
$W=F\cdot s=Fs\cos\theta$
When the force and displacement are both in the same direction, $theta = 0$. As a result, the work done would be
$W=Fs\cos0^0 = FS$
in a case when force and displacement happen to be opposite to each other $\theta = 180^0$. So, the work done would be
$W=Fs\cos180^0=-Fs$
This is the case of negative work done. The work done by frictional force when pushing a block on a hard surface is an example of negative work.

(d)

(b)

Work done by the winning team is positive because

• In a tug of war, the winning team exerts a force on the opposing team and successfully moves it over a distance.
• The losing team goes toward the winning team as the winning team tugs the rope in its direction.
• Because the applied force is in the same direction as the motion, the work done is positive.

(b)

The work done by any force that is perpendicular to the direction of displacement is equal to zero.

(d)

It is given in the question that
The blocks are uniform cubes of side =2 cm = .02 m
Mass of block $0.1 Kg$
Total work done is a change in kinetic energy plus the change in potential energy. Since the change in kinetic energy is zero. So the total work done by the child would be equal to the total change in the potential energy of blocks. 
Potential Energy $=mg\Delta h$
Potential Energy of block 1 $=0.1 \times 10 \times 0$
Potential Energy of block 2 $=0.1 \times 10 \times .02 = 0.02 J$
Potential Energy of block 3 $=0.1 \times 10 \times .02 = 0.04 J$
Thus, the total potential energy is,
$P.E. = 0+0.02+0.04=0.06 J$

(d)

we know that
$v=u+at$
$15=6+a(2)$
$a=4.5m/s^2$
$s=ut+\frac{1}{2}at^2=6(2)+0.5(4.5)(4)=21m$
$W=F\cdot s=2(4.5)(21)=189J$

(d)

$W=Fs\cos\theta$
In the case of option d $\theta=0^0$
therefore work done is maximum

(a)

Work done, $W=Fs\cos\theta$
where $F$ is the force applied, $s$ is the displacement and $\theta$ is the angle between force and displacement.
This shows that work done is independent of time.
According to our question, both forces are applied and the direction of movement of the box is the same. So we have $\theta = 0^0$
Both men are raising mass 50 kg to a height of 3 m.
Work done by first man $= mgh$
$=50\times 10\times 3$
$= 1500 J$
Work done by second man $= mgh$
$=50\times 10\times 3$
$= 1500 J$
Hence work done is in the ratio of $1500:1500=1:1$

(b)

Work done = change in kinetic energy
Therefore,
$W=\frac{1}{2}\times 0.01[(1000)^2-(500)^2]=3750J$

(a)

Distance $s=10m$ ; Force $F=5N$ ; and work done $W=25 J$
Work done is given by relation $W=Fs\cos\theta$
Therefore,
$25=5\times 10\cos\theta = 20\cos \theta$
or,
$\cos\theta=\frac{25}{50}=0.5$
or, $\theta=60^0$

(a)

When speed is constant in a circular motion, it means the work done by the centripetal force is zero.