Work, Energy and Power
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A boy is rolling a 0.5 kg ball on the frictionless floor with the speed of 20ms⁻¹. The ball gets deflected by an obstacle on the way. After deflection it moves with 5% of its initial kinetic energy. What is the speed of the ball now?
A boy is rolling a 0.5 kg ball on the frictionless floor with the speed of 20ms⁻¹. The ball gets deflected by an obstacle on the way. After deflection it moves with 5% of its initial kinetic energy. What is the speed of the ball now?
K? = ½ mv² + ½ Iω²
Since I =? mr² and v=rω,
K? = ½ mv² + ½ (? mr²) (v/r)² = ½ mv² +? mv² = (7/10)mv² = 140 J
K_f = 0.05 K?
(7/10)mv_f² = 0.05 * (140)
(7/10)mv_f² = 7
v_f² = 20
v_f = √20 ≈ 4.47 m/s
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A person pushes a box on a rough horizontal platform surface. He applies a force of 200 N over a distance of 15 m. Thereafter, he gets progressively tired and his applied force reduces linearly with distance to 100 N. The total distance through which the box has been moved is 30 m. What is the work done by the person during the total movement of the box?
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A person pushes a box on a rough horizontal platform surface. He applies a force of 200 N over a distance of 15 m. Thereafter, he gets progressively tired and his applied force reduces linearly with distance to 100 N. The total distance through which the box has been moved is 30 m. What is the work done by the person during the total movement of the box?
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Network done is equal to area under F-x curve
= 200 x 15 + ½ x 100 x 15 + 100 × 15
= 4500 + 750
= 5250
Upvote Particle A of mass m₁ moving with velocity (√3î + ĵ)ms⁻¹ collides with another particle B of mass m₂ which is at rest initially. Let V₁ and V₂ be the velocities of particles A and B after collision respectively. If m₁ = 2m₂ and after collision V₁ = (î + √3ĵ)ms⁻¹, the angle between V₁ and V₂ is:
Particle A of mass m₁ moving with velocity (√3î + ĵ)ms⁻¹ collides with another particle B of mass m₂ which is at rest initially. Let V₁ and V₂ be the velocities of particles A and B after collision respectively. If m₁ = 2m₂ and after collision V₁ = (î + √3ĵ)ms⁻¹, the angle between V₁ and V₂ is:
Blocks of masses m, 2m and 8m are arranged in a line on a frictionless floor. Another block of mass m, moving with speed v along the same line (see figure) collides with mass m in perfectly inelastic manner. All the subsequent collisions are also perfectly inelastic. By the time the last block of mass 8 m starts moving the total energy loss is p% of the original energy. Value of ' p ' is close to:
Blocks of masses m, 2m and 8m are arranged in a line on a frictionless floor. Another block of mass m, moving with speed v along the same line (see figure) collides with mass m in perfectly inelastic manner. All the subsequent collisions are also perfectly inelastic. By the time the last block of mass 8 m starts moving the total energy loss is p% of the original energy. Value of ' p ' is close to:
A Tennis ball is released from a height h and after freely falling on a wooden floor it rebounds and reaches height h/2. The velocity versus height of the ball during its motion may be represented graphically by:
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A Tennis ball is released from a height h and after freely falling on a wooden floor it rebounds and reaches height h/2. The velocity versus height of the ball during its motion may be represented graphically by:
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A particle is moving unidirectionally on a horizontal plane under the action of a constant power supplying energy source. The displacement(s) - time (t) graph that describes the motion of the particle is (graph are drawn schematically and are not to scale):
A particle is moving unidirectionally on a horizontal plane under the action of a constant power supplying energy source. The displacement(s) - time (t) graph that describes the motion of the particle is (graph are drawn schematically and are not to scale):
A cricket ball of mass 0.15 kg is thrown vertically up by a bowling machine so that it rises to a maximum height of 20 m after leaving the machine. If the part pushing the ball applies a constant force F on the ball and moves horizontally a distance of 0.2 m while launching the ball, the value of F (in N) is (g = 10 ms?²)
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A cricket ball of mass 0.15 kg is thrown vertically up by a bowling machine so that it rises to a maximum height of 20 m after leaving the machine. If the part pushing the ball applies a constant force F on the ball and moves horizontally a distance of 0.2 m while launching the ball, the value of F (in N) is (g = 10 ms?²)
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First, find the initial velocity (v) of the ball as it leaves the machine, using the kinematic equation for maximum height:
v² = u² + 2as ⇒ 0 = v² - 2gh_max
v² = 2gh_max = 2 * 10 * 20 = 400
v = 20 m/s.
Now, apply the work-energy theorem to the ball while it is being pushed by the machine. The work done
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