Question

Please answer this physics question im desperate!! 30 points and bRAINLIEST!!

Answer 1

Answer:

a) The work done can be calculated as per the equation W = F × d, where

F is the force and d is the distance ( in metres)

W = 10 × (15/100) = 1.5 Joules

b) Elastic Potential energy is a conservative force. The whole work will be stored in the elastic as Elastic Potential energy.

So the elastic Potential energy

= 1.5 Joules

c) The main assumption taken into consideration is that:

1. The whole work done is converted to Elastic Potential energy.

2. There is no other loses of energy.

d) Again, when the object is attached to the elastic, the initial kinetic energy is obtained from the stored Elastic Potential energy.

So the Kinetic energy = 1.5 joules.

e) Kinetic energy= Potential energy

=> 1/2 × m × v² = 1.5

=> 1/2 × (1/1000) × v² = 1.5

=> v² =3000

=> v = √ 3000 = 54.77 m/s.

f) The main assumption is that:

1. Whole of elastic Potential energy is converted to kinetic energy.

2. No other energy losses.

g) Using kinematics equations:

(final velocity = v = 0)

v² = u² + 2gh

=> 0 = 3000 - 2 × 10 × h

=> h = 150 metres.

h) The main assumption taken is that:

1. Force of gravity doesn't change with change in height ( remain constant).

2. The object doesn't experience air resistance.

Answer 2

Question :-

An elastic band is stretched at a distance of 15.0 cm by a force of 10.0 N

Formula used :-

$\rightarrow \sf{work \: \red{ = force \times distance}} \\ \rightarrow \sf{ \pink{kinetic \: energy }= \green{\frac{1}{2} \: m \: {v}^{2} }} \\ \rightarrow \sf{ {v}^{2} = \red{ {u}^{2} + 2ah}}$

_______________________________

Explanation :-

• Part (1) ,

• Force = 10.0 N
• Distance = 15.0 cm

therefore ,

→ Work = Force × distance

→ Work = 10 × 15 = 150 N-cm

Or ,

→ Work = 150/100 N-m or joule

→ Work = 1.5 joule

________________________________

• Part (2)

→ Total work done as elastic potential energy stored in the elastic material.

Hence,

→ Elastic potential energy is 1.5 joule.

• Part (3)

We used these assumptions ↓

• Work is stored as elastic potential energy
• No loss of energy or heat

________________________________

• Part (4)

$\star \sf{ \red{mass} \red{ =} 1.0 g \: = \pink{ \frac{1}{1000} \: kg}}$

• Velocity(v) = ?

For the given condition kinetic energy is equal to the elastic potential energy,

$\rightarrow \sf{ \red{\frac{1}{2} \: m \: {v}^{2}} = 1.5} \\ \\ \rightarrow \sf{ \pink{\frac{1}{2} \times \frac{1}{1000} {v}^{2} } = 1.5} \\ \\ \rightarrow \sf{ \green{{v}^{2} = 2000 \times 1.5}} \\ \\ \rightarrow \sf{ \red{{v}^{2} = 3000}} \\ \\ \rightarrow \sf{ \green{v \: = \sqrt{3000} } = \pink{10 \sqrt{30} } \: \: \frac{m}{s} }$

Velocity is 10√(30) m/s

______________________________

• Part (5)

• Height of object (h) = ?

• initial velocity (u) = 0

• Final velocity (v) = 10√(30) m/s

$\rightarrow \sf{ \red{{v}^{2} = } \pink{{u}^{2} + 2ah }} \\ \\ \sf{ \green{taking \: \: a \: = g}} \\ \\ \rightarrow \sf{ \pink{ {v}^{2} = {u}^{2} + 2gh}} \\ \\ \rightarrow \sf{ \red{{(10 \sqrt{30} )}^{2}} = 0 + \pink{2 \times 10 \times h}} \\ \\ \rightarrow \sf{ \green{3000 }= \red{20h}} \\ \\ \rightarrow \sf{h \: = \pink{ \frac{300}{2} }} \\ \\ \rightarrow \boxed{\sf{ \red{h }\: = \green{ 150 \: m}}}$

Height of object is 150 m

_______________________________