A 67.5 kg sprinter exerts a force of 775 N on the starting block which makes a 20 degree angle to the ground.
a) What was the horizontal acceleration of the sprinter?
b) If the force was exerted for 0.340s, with what speed did the sprinter leave the starting block?
To determine the horizontal acceleration of the sprinter, we first need to find the horizontal component of the force.
a) Horizontal component of the force: F_h = F * cos(theta)
where F = 775 N (force applied by the sprinter)
theta = 20 degrees (angle made by the force with the ground)
F_h = 775 N * cos(20 degrees)
Now, we can calculate the horizontal acceleration using Newton's second law of motion: F = m * a,
where F is the net force applied, m is the mass of the sprinter, and a is the acceleration.
a) a = F_h / m
Substituting the values:
a = (775 N * cos(20 degrees)) / 67.5 kg
To find the speed at which the sprinter leaves the starting block, we use the kinematic equation:
b) v = u + at
where v is the final velocity, u is the initial velocity (0 m/s since the sprinter starts from rest), a is the acceleration (horizontal acceleration), and t is the time period.
b) v = 0 + (a * t)
Substituting the values:
v = (775 N * cos(20 degrees) / 67.5 kg) * 0.340 s
Now, we can calculate the values for both questions:
a) Calculate the horizontal acceleration:
a = (775 N * cos(20 degrees)) / 67.5 kg
b) Calculate the speed at which the sprinter leaves the starting block:
v = (775 N * cos(20 degrees) / 67.5 kg) * 0.340 s
To find the horizontal acceleration of the sprinter, we need to resolve the force into its horizontal component. The horizontal component of the force can be found using the formula:
F_horizontal = F * cos(angle)
where F is the total force and angle is the angle the force makes with the ground.
a) Let's calculate the horizontal force:
F_horizontal = 775 N * cos(20°)
F_horizontal = 775 N * 0.9397
F_horizontal ≈ 727.32 N
Now, we can use Newton's second law of motion, which states that the force acting on an object is equal to its mass multiplied by its acceleration:
F = m * a
where F is the force, m is the mass, and a is the acceleration.
Since we have the horizontal force and the mass of the sprinter, we can solve for the horizontal acceleration:
727.32 N = 67.5 kg * a_horizontal
a_horizontal = 727.32 N / 67.5 kg
a_horizontal ≈ 10.78 m/s²
Therefore, the horizontal acceleration of the sprinter is approximately 10.78 m/s².
b) To find the speed at which the sprinter leaves the starting block, we can use the kinematic equation:
v = u + a * t
where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time.
In this case, the sprinter starts from rest (u = 0 m/s), and we have calculated the horizontal acceleration (a_horizontal) in part a. The time (t) is given as 0.340 seconds.
v = 0 + 10.78 m/s² * 0.340 s
v ≈ 3.66 m/s
Therefore, the sprinter leaves the starting block with a speed of approximately 3.66 m/s.
horizontal force component = 775 cos 20
so
a = F/m = (775/67.5)cos 220 meters/ second^2
force*time = impulse = momentum change = m v
so
775 cos 20 * 0.340 = 67.5 v