A bullet having mass 10 g is fired towards a fixed wooden block with velocity of 30 m/s. If it comes out of the block with velocity of 20 m/s, then the total work done by the resistive forces is
By Work Energy Theorem,
Work Done= Change in Kinetic Energy
⇒W= ½×(10/ 1000) × ( 20²- 30²) = -500/200= - 2.5 J
The position of a particle of mass 1 kg moving along x-axis at time t is given by x=t²/2. The work done by the force from t=0 to t=3 sec
Given, x = t²/2 differentiating v=dx/dt= t
Initial Speed Vi= 0 m/s
Final Speed Vf= 3 m/s
From Work Energy Theorem, Work done is equal to change in kinetic energy
So, Work Done = ½m (Vf² - Vi²)= 1/2 × 1× (9-0) = 4.5 J
A force act on a 2 kg particle such a way that position of the particle as a function of time is given by x = 3t -4t²+t³, Where x is in meter and t is in second. Work done during first 4 second is
Given x = 3t -4t²+t³ differentiating v = dx/dt = 3-8t+ 3t²,
Initial speed = 3 m/s
Final speed = 3-32+48= 19 m/s
From Work Energy Theorem,
Work Done = Change in Kinetic Energy= ½×2×(19²-3²)= 352 J
A 5 kg ball when falls through a height of 20 m acquires a speed of 10 m/s. Find the work done by air resistance
From work energy theorem, work done by all the forces is equal to change in kinetic energy.
So, Work done by gravity + Work done by air resistance = Final Kinetic Energy - Initial Kinetic Energy
5×10×20 + Work done by air resistance = 1/2× 5 ×(10²-0)
⇒ 1000 + Work done by air resistance = 250
⇒ Work done by air resistance = 250 -1000= -750 J
A bullet of mass 10g is fired horizontally with a velocity 1000 ms–1 from a rifle situated at a height 50m above the If the bullet reaches the ground with a velocity 500 ms–1, the work done against air resistance on the bullet is : (g = 10 ms–2)
From Work Energy Theorem,
Total Work Done = Change in Kinetic Energy
Work Done by Gravity + Work Done by Air Resistance = ½m (Vf 2 - Vi 2 )
0.001 × 10×50 + Work done by Air Resistance = ½× 0.001×( 500² - 1000²)
Solving We get, Work Done by Air Resistance = - 3755 J