**Question 1** => A spring of spring constant 50 N/m is compressed from its natural position through 1 cm. Find the work done by the spring-force on the agency compressing the spring.

**Solution** => Magnitude of **Work Done by Spring = 0.5 X Spring Constant X (Position) ^{2}**

Putting in the values we get Work Done by Spring = 0.5 X (50 N/m) X (1 cm)

^{2}

=> (25 N/m) X (1 cm

^{2}) = (25 N/m) X (0.00001 m

^{2}) =

**2.5 X 10**

^{-2}JouleSo

**Work Done by Spring = 2.5 X 10**

^{-2}Joule**Question 2** => A particle of mass 50 g is thrown vertically upwards with a speed of 20 m/s. Find the work done by the force of gravity during the time the particle goes up.

**Solution** => When particle reaches maximum height then its velocity should be zero. So **At maximum height velocity will be zero**.

As per **Third Law of Motion** => v^{2} – u^{2} = 2ah

As in this particular case, **a = -g (As particle is going opposite to Gravity Direction) and v = 0 (At Maximum Height)**

So we get 0 – u^{2} = -2gh => -u^{2} = -2gh => **h = u ^{2}/2g**

But we need to find out What Work will be done by force of gravity when this particle is thrown up, the formula for that is

**-mgh**.

**Work done by gravity = -mgh**

=> -mgh = -mg(u

^{2}/2g) = mgu

^{2}/2g = -mu

^{2}/ 2

So

**Work Done by Gravity on a particle when its thrown up = -mu**

^{2}/ 2Substituting in the values we get = -mu

^{2}/ 2 = -50 X (20)

^{2}/2 = -1.0 Joule

Final Answer =>

**Work Done by Gravity on Particle is -1.0 Joule**

**Question 3** => A block of mass m slides along a frictionless surface, what would be speed of block when it reaches at bottom of frictionless surface. (See the diagram below)

**Solution** => As per law of Conversation of energy **Total Energy of Block at Point A = Total Energy of Block**

**at Point B**(

*Equation 1*)

Total Energy of Block at Point A = Kinetic Energy of Block at Point A + Potential Energy of Block at Point A

Total Energy of Block at Point B = Kinetic Energy of Block at Point B + Potential Energy of Block at Point B

Total Energy of Block at Point A = mv

^{2}/2 + mgh = 0 + mgh = mgh

Total Energy of Block at Point B = mv

^{2}/2 + mgh = mv

^{2}/2 + 0 = mv

^{2}/2

Putting this in *Equation 1* we get => mv^{2}/2 = mgh

Solving this we get **v = (2gh) ^{1/2} = √2gh**

**Question 4** => A pendulum bob has a speed 3 m/s while passing through its lowest position. What is its speed when it makes an angle of 60° with the vertical ? The length of the pendulum is 0.5 m. Take g = 10 m/s^{2}.

**Solution** => As Per law of conservation of energy, Kinetic Energy lost in going from v0 to v1 velocity should be equal to potential energy gained as pendulum moved from height l to height lcosθ.

That’s why

mv_{o}^{2}/2 – mv_{1}^{2}/2 = mgl(1 – lcosθ)

Solving Above Formula we get **v _{1} = (v_{o}^{2} – 2gl(1 – cosθ))^{1/2}**

Putting in the values for

**v0 = 3 m/s**and

**g = 10 m/s**and

^{2}**cosθ = cos60°**

We get

**v**

_{1}= 2 ms^{-1}**Final Answer => v**

_{1}= 2 ms^{-1}**Question 5** => A porter lifts a suitcase weighing 20kg from the platform and puts it on his head which is 2.0m above the plotform. Calculate how much work is done by porter on the suitcase. (Consider g = 10 m/s^{2})

**Solution** => So when porter lifts up the suitcase all of work done by porter on suitcase would be stored in it as it’s energy. As porter is lifting suitcase away from the ground so energy stored in suitcase would be **Potential Energy**.

This means work done by porter would be equal to potential energy of suitcase at height 2.0m.**Work Done by Porter = Equal Potential Energy of Suitcase at height 2.0m**

Potential Energy of suitcase at height 2.0m = -mgh = – 20 X 10 X 2 = -392 Joule

So **Work Done by Porter = -392 Joule**