Compare vector and scalar quantities - Leaving Cert Physics - Question 6 - 2014

1. Apparatus and Arrangement: Use weights and pulleys to set up the experiment. 2. Procedure and Measurements: Adjust and read each vector force using a force sensor or spring scale. 3. Observation and Result: Measure the angle and find the resultant using vector addition, observing the direction and magnitude.

To find the net force, we need to consider the horizontal and vertical components of the forces:

Horizontal force applied by the golfer:

$F_{horizontal} = 277N \cos(24.53°) = 252N$

Vertical force applied by golfer:

$F_{vertical} = 277N \sin(24.53°) = 115N$

The net force, considering friction:

$Net \, Force = F_{horizontal} - F_{friction} = 252N - 252N = 0N$

The net force being zero indicates that the golfer is moving at a constant velocity. There is no acceleration, and thus the forces are balanced.

Newton's second law states that the force (F) acting on an object is equal to the mass (m) of the object multiplied by the acceleration (a) of the object:

$F = ma$ Here, 'k' is defined as 1 N, where 1 Newton is the force needed to accelerate 1 kg at a rate of 1 m/s².

Using the formula:

$F = \frac{mv}{t} \Rightarrow v = \frac{Ft}{m}$

Substituting the values:

$F = 5.3 kN = 5300 N, \quad t = 0.54 ms = 0.00054 s, \quad m = 45 g = 0.045 kg$

The calculation yields:

$v = \frac{5300 N \times 0.00054 s}{0.045 kg} = 63.6 m/s$

First, calculate the vertical component of the initial velocity:

$v_y = v \sin(15°) = 63.6 m/s \sin(15°) = 16.6 m/s$

Using the formula for the maximum height:

$h = \frac{v_y^2}{2g}$

Substituting:

$h = \frac{(16.6 m/s)^2}{2 \times 9.8 m/s^2} = 13.82 m$