Work and Energy Questions

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 cm2) = (25 N/m) X (0.00001 m2) = 2.5 X 10-2 Joule
So 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 => v2 – u2 = 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 – u2 = -2gh => -u2 = -2gh => h = u2/2g
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(u2/2g) = mgu2/2g = -mu2 / 2
So Work Done by Gravity on a particle when its thrown up = -mu2 / 2
Substituting in the values we get = -mu2 / 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)

block of mass m sliding down a frictionless surface

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 = mv2/2 + mgh = 0 + mgh = mgh
Total Energy of Block at Point B = mv2/2 + mgh = mv2/2 + 0 = mv2/2

Putting this in Equation 1 we get => mv2/2 = 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/s2.

Swinging Pendulum

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
mvo2/2 – mv12/2 = mgl(1 – lcosθ)
Solving Above Formula we get v1 = (vo2 – 2gl(1 – cosθ))1/2
Putting in the values for v0 = 3 m/s and g = 10 m/s2 and cosθ = cos60°
We get v1 = 2 ms-1
Final Answer => v1 = 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/s2)

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


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