Advertisements
Advertisements
प्रश्न
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.
Advertisements
उत्तर
The specifications given by the company are:
\[\text{ Mass, m = 95 kg } \]
\[\text{ Maximum power, P_m = 3 . 5 hp } \]
\[\text{ Maximum speed, v}_\text{m }= 60 \text{ km /h} \]
\[ = \frac{50}{3} \text{ m/s } \]
\[\text{ Pick up time to get maximum speed, t}_\text{m }= 5 \sec\]
So, the maximum acceleration that can be produced,
\[\text{ a }= \frac{50}{3 \times 5} = \frac{10}{3} \text{ m/ s} ^2\]
So, the driving force,
\[\text{ F = ma } = 95 \times \left( \frac{10}{3} \right)\]
\[ = \frac{950}{3} \text{ N } \]
\[\text{ Max speed, } \nu = \frac{\text{ p } }{\text{ F } }\]
\[ \Rightarrow \text{ v }=3.5\times746\times\frac{3}{950}\Rightarrow8.2 \text{ m/s } \]
As the scooter can reach a maximum of 8.2 m/s while producing a force of 950/3 N, the specifications given are not correct.
APPEARS IN
संबंधित प्रश्न
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. Calculate the kinetic energy of Griffith-Joyner at her full speed.
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?
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?
A small block of mass 200 g is kept at the top of a frictionless incline which is 10 m long and 3⋅2 m high. How much work was required (a) to lift the block from the ground and put it an the top, (b) to slide the block up the incline? What will be the speed of the block when it reaches the ground if (c) it falls off the incline and drops vertically to the ground (d) it slides down the incline? Take g = 10 m/s2.
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.
The bob of a pendulum at rest is given a sharp hit to impart a horizontal velocity \[\sqrt{10 \text{ gl }}\], where l is the length of the pendulum. Find the tension in the string when (a) the string is horizontal, (b) the bob is at its highest point and (c) the string makes an angle of 60° with the upward vertical.
A simple pendulum consists of a 50 cm long string connected to a 100 g ball. The ball is pulled aside so that the string makes an angle of 37° with the vertical and is then released. Find the tension in the string when the bob is at its lowest position.
A simple pendulum of length L with a bob of mass m is deflected from its rest position by an angle θ and released (following figure). The string hits a peg which is fixed at a distance x below the point of suspension and the bob starts going in a circle centred at the peg. (a) Assuming that initially the bob has a height less than the peg, show that the maximum height reached by the bob equals its initial height. (b) If the pendulum is released with \[\theta = 90^\circ \text{ and x = L}/2\] , find the maximum height reached by the bob above its lowest position before the string becomes slack. (c) Find the minimum value of x/L for which the bob goes in a complete circle about the peg when the pendulum is released from \[\theta = 90^\circ \]
A particle of mass m is kept on the top of a smooth sphere of radius R. It is given a sharp impulse which imparts it a horizontal speed ν. (a) Find the normal force between the sphere and the particle just after the impulse. (b) What should be the minimum value of ν for which the particle does not slip on the sphere? (c) Assuming the velocity ν to be half the minimum calculated in part, (b) find the angle made by the radius through the particle with the vertical when it leaves the sphere.
Figure ( following ) shows a smooth track which consists of a straight inclined part of length l joining smoothly with the circular part. A particle of mass m is projected up the incline from its bottom. Find the minimum projection-speed \[\nu_0\] for which the particle reaches the top of the track.
Figure ( following ) shows a smooth track which consists of a straight inclined part of length l joining smoothly with the circular part. A particle of mass m is projected up the incline from its bottom. Assuming that the projection-speed is \[\nu_0\] and that the block does not lose contact with the track before reaching its top, find the force acting on it when it reaches the top.
An electron and a proton are moving under the influence of mutual forces. In calculating the change in the kinetic energy of the system during motion, one ignores the magnetic force of one on another. This is because ______.
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 bullet of mass m fired at 30° to the horizontal leaves the barrel of the gun with a velocity v. The bullet hits a soft target at a height h above the ground while it is moving downward and emerges out with half the kinetic energy it had before hitting the target.
Which of the following statements are correct in respect of bullet after it emerges out of the target?
- The velocity of the bullet will be reduced to half its initial value.
- The velocity of the bullet will be more than half of its earlier velocity.
- The bullet will continue to move along the same parabolic path.
- The bullet will move in a different parabolic path.
- The bullet will fall vertically downward after hitting the target.
- The internal energy of the particles of the target will increase.
Give example of a situation in which an applied force does not result in a change in kinetic energy.
Two bodies of unequal mass are moving in the same direction with equal kinetic energy. The two bodies are brought to rest by applying retarding force of same magnitude. How would the distance moved by them before coming to rest compare?
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.
