At the instant a race began, a 65 kg sprinter exerted a force of 720 N on the starting block at a 22° angle with respect to the ground. (a) What was the horizontal acceleration of the sprinter? (b) If the force was exerted for 0.32 s, with what speed did the sprinter leave the starting block?

The acceleration of an object can be obtained by using the equation of motion. It depends on the value of the applied force. Hence, as the value of the force increases, the value of the acceleration also increases.

Given data:

The mass of the sprinter is $m=65\mathrm{kg}$.

The applied force is $F=720\mathrm{N}$.

The angle is $\theta =22°$.

The time is $t=0.32s$.

(a)

Draw a free-body diagram.

Here, $F\cos \theta$is the horizontal component of the force, $F\sin \theta$is the vertical component of the force, $N$is the normal force, and $mg$is the weight of the sprinter.

Applying the equilibrium condition along the horizontal direction,

$\begin{array}{c}{F}_x=F\cos \theta \\ m{a}_x=F\cos \theta \\ a_x=\frac{F\cos \theta }{m}\end{array}$

Substituting the known values in the above equation,

$\begin{array}{c}{a}_x=\frac{\left(720\mathrm{N}\right)\cos \left(22°\right)}{\left(65\mathrm{kg}\right)}\\ a_x=10.27\mathrm{m}{/\mathrm{s}}^2\end{array}$

Thus, the horizontal acceleration of the sprinter is$10.27\mathrm{m}{/\mathrm{s}}^2$.

The final speed of the sprinter can be calculated as

$v=u+a_{x}t$.

Here,$u=0\mathrm{m}/\mathrm{s}$is the initial velocity of the sprinter.

Substituting the values in the above expression,

$\begin{array}{c}v=\left(0\mathrm{m}/\mathrm{s}\right)+\left(10.27\mathrm{m}{/\mathrm{s}}^2\right)\left(0.32s\right)\\ v=3.28\mathrm{m}/\mathrm{s}\end{array}$

Thus, the final speed of the sprinter is$3.28\mathrm{m}/\mathrm{s}$.