At the instant a race began, a 60-kg sprinter exerted a force of 669-N on the starting block at a 24 angle with respect to the ground. If this force was exerted for 0.44-s, with what speed (in m/s) did the sprinter leave the starting block?
Fx = 669 cos 24
change in x momentum or impulse = Fx t
so
669 cos 24 * .44 = 60 v
To find the speed with which the sprinter leaves the starting block, we need to apply Newton's second law of motion, which states that the net force applied to an object is equal to the product of its mass and acceleration. In this case, the sprinter exerts a force on the starting block, and the reaction force from the block propels the sprinter forward.
First, we need to resolve the force into its vertical (F_vertical) and horizontal (F_horizontal) components using trigonometry.
F_vertical = F * sin(theta)
F_horizontal = F * cos(theta)
where F is the magnitude of the force applied (669 N) and theta is the angle with respect to the ground (24 degrees).
F_vertical = 669 N * sin(24) = 669 N * 0.3953 ≈ 264.51 N
F_horizontal = 669 N * cos(24) = 669 N * 0.9187 ≈ 613.68 N
The vertical component of the force does not contribute to the sprinter's horizontal acceleration since it acts perpendicular to the sprinter's motion. Thus, we only need to consider the horizontal component.
Next, we can calculate the acceleration (a) of the sprinter using Newton's second law:
F_horizontal = m * a
where m is the mass of the sprinter (60 kg). Rearranging the equation, we have:
a = F_horizontal / m = 613.68 N / 60 kg ≈ 10.23 m/s^2
Finally, we can use the equation of motion to find the speed (v) of the sprinter when leaving the starting block:
v = u + a * t
where u is the initial velocity (which is zero in this case), a is the acceleration, and t is the time interval for which the force is applied.
Substituting the known values:
v = 0 + (10.23 m/s^2) * 0.44 s ≈ 4.51 m/s
Therefore, the sprinter leaves the starting block with a speed of approximately 4.51 m/s.