Question 1:
moderateA 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,Â
Work Done= Change in Kinetic Energy
⇒W= ½×(10/ 1000) × ( 20²- 30²) = -500/200= - 2.5 J
Question 2:
moderateThe 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= tÂ
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
Question 3:
moderateA 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 isÂ
Given x = 3t -4t²+t³ differentiating v = dx/dt = 3-8t+ 3t²,Â
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 JÂ
Question 4:
moderateA 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
Question 5:
moderateA 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
Question 6:
moderateThe only force acting on a \(2\text{ kg}\) body as it moves along the positive x axis has component \(F_x = -6x\text{ N}\), where x is in metre. The velocity of the body at \(x = 3\text{ m}\) is \(8\text{ m/s}\). The velocity of the body at \(x = 4\text{ m}\) is:
Applying the work-energy theorem, \(W = \Delta K ⇒ \int_{3}^{4} -6x , dx = \frac{1}{2} m(v_f^2 - v_i^2)\). Integrating gives \(-3(16 - 9) = \frac{1}{2} (2) (v_f^2 - 64)⇒ -21 = v_f^2 - 64\), which yields \(v_f = \sqrt{43} \approx 6.6\text{ m/s}\).
Question 7:
moderateA force \(F = (-0.6x^2 + 2.5)\text{N}\) displaces a particle of mass 2 kg from \(x = 0\) to \(x = 10\text{ m}\). If velocity of particle at \(x = 0\) was 20 m/s, then velocity at \(x = 10\text{ m}\) will be
Work done \(W = \int_{0}^{10} (-0.6x^2 + 2.5)dx = [-0.2x^3 + 2.5x]_0^{10} = -175\text{ J}\). From the work-energy theorem, \(W = \frac{1}{2}m(v_f^2 - v_i^2) \implies -175 = \frac{1}{2}(2)(v_f^2 - 400)\text{, giving } v_f = 15\text{ m/s}\).
Question 8:
moderateA body of mass $1\text{ kg}$ is thrown upwards with a velocity $20\text{ m/s}$. It momentarily comes to rest after attaining a height of $18\text{ m}$. How much energy is lost due to air friction? ($g = 10\text{ m/s}^2$)
(2009)
Apply conservation of energy with non-conservative forces: initial energy $E_i = \frac{1}{2}mv^2 = 200\text{ J}$. Final potential energy at $18\text{ m}$ is $U_f = mgh = 180\text{ J}$. Energy lost against air friction is $\Delta E = E_i - U_f = 200 - 180 = 20\text{ J}$.
Question 9:
moderateA body of mass $3\text{ kg}$ is under a constant force which causes a displacement $s$ in meters in it, given by the relation $s = \frac{1}{3} t^2$, where $t$ is in $s$. Work done by the force in $2\text{ s}$ is:
(2006)
Find acceleration by differentiating displacement twice: $v = \frac{2}{3}t$ and $a = \frac{2}{3}\text{ m/s}^2$. Force is $F = ma = 2\text{ N}$. At $t = 2\text{ s}$, displacement is $s = \frac{4}{3}\text{ m}$. Work done $W = F \times s = \frac{8}{3}\text{ J}$.
Question 10:
easyIf $x = 3 – 4t^2 + t^3$, then work done in first $4\text{ s}$ will be (Mass of the particle is $3\text{ gram}$):
(1998)
Use work-energy theorem $W = \Delta K = \frac{1}{2}m(v_f^2 - v_i^2)$. Velocity is $v = -8t + 3t^2$. At $t = 0$, $v_i = 0$ and at $t = 4\text{ s}$, $v_f = 16\text{ m/s}$. Substituting values with $m = 3\text{ g} = 0.003\text{ kg}$ gives $W = 384\text{ mJ}$.