a shot putter launches a 7.2kg shot by pushing it along straight line of length 1.6 m and angle of 34.10deg FROM HORIZONTAL ACC. THE SHOT FROM LAUNCH SPEED TO INITIAL SPEED OF 2.5M/S .THE SHOT LEAVES THE HAND AT HEIGHT OF 2.11M AND ANGLE OF34.10DEG AND LANDS AT THE DISTANCE OF 15.90M WHAT IS THE MAGNITUDE OF ATHLETE FORCE ON THE SHOT DURING ACC. PHASE???

HINT-;-TREAT THE MOTION THROUGH ACC PHASE AS THOUGH IT WAS ALONG THE RAMP AT GIVEN ANGLE

To determine the magnitude of the athlete's force on the shot during the acceleration phase, we can break down the problem into manageable steps. We'll analyze the forces acting on the shot put while it is being accelerated along the ramp, taking into account the angle of launch and the weight of the shot. Let's go through this step by step.

Understanding the Forces Involved

When the shot put is being accelerated, two main forces act on it: the gravitational force and the force applied by the athlete. The gravitational force can be calculated using the formula:

• Weight (W) = mass (m) × gravitational acceleration (g)

Here, the mass of the shot put (m) is 7.2 kg, and the gravitational acceleration (g) is approximately 9.81 m/s². Thus, the weight of the shot put is:

W = 7.2 kg × 9.81 m/s² = 70.71 N

Components of Forces

Next, we need to resolve the weight of the shot put into components parallel and perpendicular to the ramp. The angle of the ramp is given as 34.10 degrees. The components can be calculated as follows:

• Weight parallel to the ramp (W_parallel) = W × sin(θ)
• Weight perpendicular to the ramp (W_{perpendicular}) = W × cos(θ)

Calculating these components:

W_parallel = 70.71 N × sin(34.10°) ≈ 70.71 N × 0.5748 ≈ 40.63 N

W_{perpendicular} = 70.71 N × cos(34.10°) ≈ 70.71 N × 0.8192 ≈ 57.83 N

Net Force and Acceleration

Now, we can find the net force acting on the shot put during the acceleration phase. The net force (F_net) is the difference between the force exerted by the athlete (F_applied) and the weight component acting parallel to the ramp:

F_net = F_applied - W_parallel

Using Newton's second law, we can express the net force in terms of mass and acceleration:

F_net = m × a

To find the acceleration (a), we can use the kinematic equation:

v² = u² + 2as

Where:

• v = final velocity (2.5 m/s)
• u = initial velocity (0 m/s)
• s = distance (1.6 m)

Rearranging the equation to solve for acceleration:

a = (v² - u²) / (2s) = (2.5² - 0²) / (2 × 1.6) = 6.25 / 3.2 ≈ 1.95 m/s²

Calculating the Applied Force

Now that we have the acceleration, we can substitute it back into the net force equation:

F_net = m × a = 7.2 kg × 1.95 m/s² ≈ 14.04 N

Now we can find the applied force:

F_applied = F_net + W_parallel = 14.04 N + 40.63 N ≈ 54.67 N

Final Result

The magnitude of the athlete's force on the shot during the acceleration phase is approximately 54.67 N. This value represents the total force exerted by the athlete to overcome both the gravitational component acting along the ramp and to provide the necessary acceleration to the shot put.