Question

An iron ball of mass 9 kg is placed at one end of the lever at a distance of 5m from point o (fulcrum). The length of lever is 20m . The effort is to be given at other end to lift the ball is ?

Answer 1

Answer:

please mark as brainlest answer

I can do an approximate calculation which will explain why he hits the ball farther with a faster pitch. I will assume that the collision between the ball and the bat is perfectly elastic (no energy is lost) which, while not exactly true, should be good enough for our purposes. The equation to use is

vball=[uball(mball-mbat)+2mbatubat]/(mball+mbat).

Here, v is the velocity after the two collide and u is the velocity before the collision and m is the mass; I hope it is obvious what the subscripts denote. Now, I looked up reasonable values of masses, incoming velocities, and bat speeds for high school kids:

mball=0.145 kg

mbat=1 kg

ubat=9 m/s (about 31 mph)

uball=-30 m/s (about 70 mph) for the game pitcher

uball=-13 m/s (about 30 mph) for warmup tosses

Note that I have chosen the direction from the plate to the pitcher to be positive, hence the negative signs on the pitched velocities. Now, I find that vball=38.1 m/s for the pitched ball, vball=25.4 m/s for the tossed ball. All other things being equal, the faster the pitched ball, the faster the hit ball. All this is for a one-dimensional calculation (all velocities right before and right after the collision are along a single straight line) and ignore the fact that there is the force which the batter is exerting on the bat; so do not expect quantitative agreement with real-life situations. Nevertheless, if all the possible complications could be included, the qualitative result would be unchanged.

Your second question I cannot answer because body mechanics are far too complex to deal with using simple physics. This is the domain for a good, experienced hitting coach who knows proper batting techniques from years of experience. I couldn't help noticing, though, that Babe Ruth had a low-hands swing! One thing in your question bothers me, though: "…the force produced after contact…"? Apart from gravity and air drag, there are no forces on the ball after it leaves the bat. Your son affects the ball only during the time the ball and bat are in contact with each other.

Answer 2

The effort at other end to lift the ball is 2.25 kg.

Explanation:

Given that,

Mass of ball = 9 kg

Distance from fulcrum = 5 m

Lever length = 20 m

We need to calculate the effort at other end to lift the ball

Using formula of moment

moments of mass= moments of effort

$M\times d=effort\times d'$

$effort=\dfrac{M\times d}{d'}$

Where, M = mass of ball

d = distance from fulcrum

d' = length of leaver

Put the value into the formula

$effort=\dfrac{9\times5}{20}$

$effort =2.25\ kg$

Hence, The effort at other end to lift the ball is 2.25 kg.

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Topic : moments of effort

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