Advertisements
Advertisements
प्रश्न
An unruly demonstrator lifts a stone of mass 200 g from the ground and throws it at his opponent. At the time of projection, the stone is 150 cm above the ground and has a speed of 3 m/s. Calculate the work done by the demonstrator during the process. If it takes one second for the demonstrator to lift the stone and throw it, what horsepower does he use?
Advertisements
उत्तर
\[\text{ Given } , \]
\[\text{ Mass of the stone, m = 200 g = 0 . 2 kg } \]
\[\text{ Heightto which the stoneis lifted, h = 150 cm = 1 . 5 m } \]
\[\text{ Velocity of the projection, } \nu = 3 \text{ m/s } \]
\[\text{ Time, t = 1 s } \]
\[\text{ Total work done, W = K . E . + P . E . } \]
\[\text{ W } = \frac{1}{2}\text{ m } \nu^2 + \text{ mgh } \]
\[ = \left( \frac{1}{2} \right) \times \left( 0 . 2 \right) \times 9 + \left( 0 . 2 \right) \left( 9 . 8 \right) \times \left( 1 . 5 \right)\]
\[ = 3 . 84 \text{ J } \]
1 hp = 764 watt
Horsepower used by demonstrator
\[= \frac{3 . 84}{746} = \left( 5 . 14 \right) \times {10}^{- 3}\]
Therefore, power used by the demonstrator to lift and throw the stone is 5.14 × 10-3 hp.
APPEARS IN
संबंधित प्रश्न
In figure (i) the man walks 2 m carrying a mass of 15 kg on his hands. In Figure (ii), he walks the same distance pulling the rope behind him. The rope goes over a pulley, and a mass of 15 kg hangs at its other end. In which case is the work done greater?
Is work-energy theorem valid in non-inertial frames?
A ball is given a speed v on a rough horizontal surface. The ball travels through a distance l on the surface and stops. what are the initial and final kinetic energies of the ball?
Consider the situation of the previous question from a frame moving with a speed v0 parallel to the initial velocity of the block. (a) What are the initial and final kinetic energies? (b) What is the work done by the kinetic friction?
The US athlete Florence Griffith-Joyner won the 100 m sprint gold medal at Seoul Olympics in 1988, setting a new Olympic record of 10⋅54 s. Assume that she achieved her maximum speed in a very short time and then ran the race with that speed till she crossed the line. Take her mass to be 50 kg. What power Griffith-Joyner had to exert to maintain uniform speed?
A water pump lifts water from 10 m below the ground. Water is pumped at a rate of 30 kg/minute with negligible velocity. Calculate the minimum horsepower that the engine should have to do this.
A scooter company gives the following specifications about its product:
Weight of the scooter − 95 kg
Maximum speed − 60 km/h
Maximum engine power − 3⋅5 hp
Pick up time to get the maximum speed − 5 s
Check the validity of these specifications.
Consider the situation shown in the following figure. The system is released from rest and the block of mass 1 kg is found to have a speed 0⋅3 m/s after it has descended a distance of 1 m. Find the coefficient of kinetic friction between the block and the table.
A block of mass 5 kg is suspended from the end of a vertical spring which is stretched by 10 cm under the load of the block. The block is given a sharp impulse from below, so that it acquires an upward speed of 2 m/s. How high will it rise? Take g = 10 m/s2.
A block of mass 250 g is kept on a vertical spring of spring constant 100 N/m fixed from below. The spring is now compressed 10 cm shorter than its natural length and the system is released from this position. How high does the block rise ? Take g = 10 m/s2.
A chain of length l and mass m lies on the surface of a smooth sphere of radius R > l with one end tied to the top of the sphere. Find the tangential acceleration \[\frac{d\nu}{dt}\] of the chain when the chain starts sliding down.
A smooth sphere of radius R is made to translate in a straight line with a constant acceleration a. A particle kept on the top of the sphere is released at zero velocity with respect to the sphere. Find the speed of the particle with respect to the sphere as a function of the angle θ it slides.
A man, of mass m, standing at the bottom of the staircase, of height L climbs it and stands at its top.
- Work done by all forces on man is equal to the rise in potential energy mgL.
- Work done by all forces on man is zero.
- Work done by the gravitational force on man is mgL.
- The reaction force from a step does not do work because the point of application of the force does not move while the force exists.
A raindrop of mass 1.00 g falling from a height of 1 km hits the ground with a speed of 50 ms–1. Calculate
- the loss of P.E. of the drop.
- the gain in K.E. of the drop.
- Is the gain in K.E. equal to a loss of P.E.? If not why.
Take g = 10 ms–2
Suppose the average mass of raindrops is 3.0 × 10–5 kg and their average terminal velocity 9 ms–1. Calculate the energy transferred by rain to each square metre of the surface at a place which receives 100 cm of rain in a year.
A rocket accelerates straight up by ejecting gas downwards. In a small time interval ∆t, it ejects a gas of mass ∆m at a relative speed u. Calculate KE of the entire system at t + ∆t and t and show that the device that ejects gas does work = `(1/2)∆m u^2` in this time interval (neglect gravity).
