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A shot putter launches a 7.260 kgshot by pushing it along a straight line of length 1.650 mand at an angle of34.10from the horizontal, accelerating the shot to the launch speed from its initial speed of 2.500 m/s(which is due to the athlete鈥檚 preliminary motion).The shot leaves the hand at a height of 2.110 mand at an angle of34.10and it lands at a horizontal distance of 15.90 m. What is the magnitude of the athlete鈥檚 average force on the shot during the acceleration phase? (Hint:Treat the motion during the acceleration phase as though it were along a ramp at the given angle.)

Short Answer

Expert verified

The magnitude of the athlete鈥檚 average force on the shot during the acceleration phase is 334.8 N .

Step by step solution

01

Given oinformation

It is given that,

The mass of the shot isM=7.26kg

The distance travelled by the shotx=1.65m

The angle at which the shot is launched,=34.10

The initial velocity of the shot v0=2.5m/s

Shot leaves the hand at a height y0=2.110m

02

Determining the concept

This problem deals with projectile path of an object. In projectile motion the path traced by the object is called trajectory. Also, this problem involves Newton鈥檚 Second law of motion. Using the Newton鈥檚 second law and kinematic equation of motion, the magnitude of the athlete鈥檚 average force on the shot during the acceleration phase can be calculated.

Formulae:

The equation for projectile path is,

y=y0+xtan0-gx22V0cos02 (i)

The Newton鈥檚 second law is,

Fnet=Ma (ii)

where, Fnet is the net force, Mis mass and a is an acceleration.

The velocity in Newton鈥檚 third kinematic equation is given by,

vf2=v02+2ax (iii)

where,vf is the final velocity,v0 is the initial velocity, a is an acceleration andx is displacement.

03

Determining the magnitude of the athlete’s average force

With the given values the equation (i) can be written as,

0-2.11=15.9tan34.10-9.815.922v0cos34.102-2.11=10.769-2477.542v020.686v02=140.3v0=11.85

Using equation (iii) the acceleration can eb written as,

a=vf2-v022xa=11.852-2.52/21.65a=40.63m/s2

Therefore, the magnitude of the athlete鈥檚 average force on the shot during the acceleration phase is,

F=Ma+MgsinF=7.2640.63+9.8sin34.10F=334.8N

Therefore, the magnitude of the athlete鈥檚 average force on the shot during the acceleration phase is 334.8 N .

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