Question

1. The velocity of a body moving in a straight line is increased by applying a
constant force F, for some distance in the direction of the motion. Prove
that the increase in the kinetic energy of the body is equal to the work done
by the force on the body.

2. Is it possible that an object is in the state of accelerated motion due to
external force acting on it, but no work is being done by the force. Explain
it with an example.

3. A ball is dropped from a height of 10 m. If the energy of the ball reduces by 40% after

striking the ground, how much high can the ball bounce back? (g = 10 m s–²)

4. If an electric iron of 1200 W is used for 30 minutes everyday, find electric

energy consumed in the month of April.

Explaination needed. ​

Answer 1

Explanation:

The velocity of a body moving in a straight line is increased by applying a constant force F, for some distance in the direction of the motion. ... kinetic energy of the body is equal to the work done by the force on the body

Answer 2

$\large\color{purple}\underline{\underline{{ \boxed {Answer 01 } }}}$

➡ We know that,

$\tt \: v²-u² = 2 as \\ \tt \: s = \frac {(v²-u²)}{2a} \\ \\ \tt \: F = ma \\ \tt \:So, Work \: done (W) =F × s\\ \\ \tt \:=W = ma × s\\ \\ \tt \:=W = ma × \frac{(v²-u²)}{2a} \\ \\ \tt \:=W = \frac{ (mv²-mu²)}{2}\\ \\ \tt \:= W = \frac{1}{2} mv² - \frac{1}{2} mu²$

So, the increase in kinetic energy of a body is 1/2 mv² - 1/2 mu² which is equal to work done by the force on the body.

Hence proved.

$\large\color{purple}\underline{\underline{{ \boxed {Answer 02 } }}}$

We know that

$\tt \: Displacement = final \: position-initial \: position$

In this case displacement is zero

so;

$\tt \: work \: done =force × displacement$

=force× 0

=o (zero)

so work done is zero in case of vertically projected body on earth.

$\large\color{purple}\underline{\underline{{ \boxed {Answer 03 } }}}$

Height at which the ball will bounce back is 6 meters.

Given:

• Ball dropped at the height (h) = 10 m
• g = 10ms

Total energy of the ball = m × g× h

= m × 10 × 10

= 100 mJ

The ball loses 40 percent of its initial energy on striking the ground level

$\tt \: ⇒ ( \frac{40}{100} ) × 100 \: mJ$

The ball loses 40 percent of its initial energy on striking the ground level = 40 mJ

Energy left to hit the ground = 100 m J – 40 m J

= 60 mJ

∴ Final energy of the ball = 60 mJ

Height at which the ball will bounce back

H = 60/10

H = 6m

$\large\color{purple}\underline{\underline{{ \boxed {Answer 04 } }}}$

Electric energy produced in the month of April is 18 kWh

Given,

• P = 1200 W = 1.2 kW      
• Time = 30 minutes = 1 / 2 hrs

We know,

No. of days in April = 30 days

Total time iron being used

$\tt \: T = \frac{1}{2} \times 30 = 15 hrs$

We know,

$\tt \: Energy \: consumed = Pt \\ \tt \: Substituting \: the \: value \: of \: P \: and \: T \\ \tt \: = 1.2 kW \times 15 hr \\ \tt \: = 18 kWh$

Hence,

Electric energy produced in the month of April is 18 kWh

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$\large\color{green}\underline{\underline{{ \boxed { \tt {{@WildCat7083} }}} }}$