An athlete with a mass of 62 kg jumps and lands on the ground on his feet. Th e ground exerts a total force of 1.1 x 10 ^ 3 N [backward 55°up] on his feet. Calculate the acceleration of the athlete.
The acceleration of the athlete can be calculated using the equation F = ma, where F is the force, m is the mass, and a is the acceleration.
a = F/m
a = (1.1 x 10^3 N) / (62 kg)
a = 17.74 m/s^2
Well, well, well, we have an athlete in action, huh? Let's see what's going on here.
To calculate the acceleration of our athletic friend, we need to use Newton's second law of motion, which states that force equals mass times acceleration (F = m * a). But since we already have the force (1.1 x 10^3 N) and the mass (62 kg), we can rearrange the equation to solve for acceleration:
a = F / m
Plugging in the values, we get:
a = (1.1 x 10^3 N) / (62 kg)
Now, let's divide these numbers and see what we get:
a ≈ 17.74 m/s²
Voila! The acceleration of our mighty athlete is approximately 17.74 m/s². Now, let's hope he sticks the landing with style and grace!
To calculate the acceleration of the athlete, we need to use Newton's second law of motion, which states that the acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass. The formula is given by:
a = F_net / m
Where:
a = acceleration
F_net = net force
m = mass
In this case, the net force exerted on the athlete by the ground is given as 1.1 x 10^3 N [backward 55° up]. To calculate the horizontal component of the force, we can use the sine function:
F_horizontal = F_net * sin(55°)
Now, substituting the known values into the formula, we have:
F_horizontal = 1.1 x 10^3 N * sin(55°)
Next, we can calculate the acceleration:
a = F_horizontal / m
Substituting the known values, we obtain:
a = (1.1 x 10^3 N * sin(55°)) / 62 kg
By evaluating the expression, we can determine the acceleration of the athlete.
To calculate the acceleration of the athlete, we can use Newton's second law, which states that the net force is equal to the product of mass and acceleration:
Fnet = m * a
In this case, the net force is the force exerted by the ground, which is given as 1.1 x 10^3 N [backward 55°up]. We need to break this force into its horizontal and vertical components.
The horizontal component is given by Fx = F * cos(θ) = 1.1 x 10^3 N * cos(55°) = 1.1 x 10^3 N * 0.5736 ≈ 632.96 N
The vertical component is given by Fy = F * sin(θ) = 1.1 x 10^3 N * sin(55°) = 1.1 x 10^3 N * 0.8192 ≈ 901.12 N
Since we are interested in calculating the acceleration, which is a vector quantity, we need to calculate both the horizontal and vertical components of acceleration.
The horizontal acceleration (ax) can be calculated using the horizontal component of the net force:
Fnetx = m * ax
So, ax = Fnetx / m = 632.96 N / 62 kg ≈ 10.2 m/s^2 (rounded to one decimal place)
The vertical acceleration (ay) can be calculated using the vertical component of the net force:
Fnety = m * ay
So, ay = Fnety / m = 901.12 N / 62 kg ≈ 14.5 m/s^2 (rounded to one decimal place)
Therefore, the acceleration of the athlete is approximately 10.2 m/s^2 horizontally (backward) and 14.5 m/s^2 vertically (upward).