Question

A shot putter launches a 7.260 kgshot by pushing it along a straight line of length 1.650 mand at an angle of$\mathbf{34}\mathbf{.}\mathbf{10}\mathbf{°}$from the horizontal, accelerating the shot to the launch speed from its initial speed of 2.500 m/s(which is due to the athlete’s preliminary motion).The shot leaves the hand at a height of 2.110 mand at an angle of$\mathbf{34}\mathbf{.}\mathbf{10}\mathbf{°}$and it lands at a horizontal distance of 15.90 m. What is the magnitude of the athlete’s 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

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

Step-by-step solution

Given oinformation

It is given that,

The mass of the shot is$M=7.26\mathrm{kg}$

The distance travelled by the shot$x=1.65\mathrm{m}$

The angle at which the shot is launched,$\theta =34.10°$

The initial velocity of the shot $v_0=2.5\mathrm{m}/\mathrm{s}$

Shot leaves the hand at a height $y_0=2.110\mathrm{m}$

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’s Second law of motion. Using the Newton’s second law and kinematic equation of motion, the magnitude of the athlete’s average force on the shot during the acceleration phase can be calculated.

Formulae:

The equation for projectile path is,

$y=y_0+x\tan {\theta }_0-\frac{g{x}^2}{2{\left(V_0\cos {\theta }_0\right)}^2}$ (i)

The Newton’s second law is,

${\mathrm{F}}_{\mathrm{net}}=\sum _{}\mathrm{Ma}$ (ii)

where, $F_{net}$ is the net force, Mis mass and a is an acceleration.

The velocity in Newton’s third kinematic equation is given by,

$v_f^2=v_0^2+2a\left(∆x\right)$ (iii)

where,$v_f$ is the final velocity,$v_0$ is the initial velocity, a is an acceleration and$∆x$ is displacement.

Determining the magnitude of the athlete’s average force

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

$$\begin{gathered} 0-2.11=15.9\tan \left(34.10\right)-\frac{\left(9.8\right){\left(15.9\right)}^2}{2{\left(v_0\cos \left(34.10\right)\right)}^2} \\ -2.11=10.769-\frac{2477.54}{2v_0^2\left(0.686\right)} \\ {v}_0^2=140.3 \\ {v}_0=11.85 \end{gathered}$$

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

$$\begin{gathered} a=\frac{v_f^2-v_0^2}{2\left(∆x\right)} \\ a=\left[{\left(11.85\right)}^2-{\left(2.5\right)}^2\right]/\left[2\left(1.65\right)\right] \\ a=40.63\mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$

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

$$\begin{gathered} F=Ma+Mg\sin \theta \\ F=7.26\left(40.63+9.8\times \sin 34.10°\right) \\ F=334.8\mathrm{N} \end{gathered}$$

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