Question

1. an athlete with a mass of 75 kg is setting up for a race. starting from rest, the athlete reaches a top speed of 33 km/h e in a total of 15 metres e. assume there is no air resistance. what is the net force on the athlete? note: divide velocity in km/h by 3.6 to get velocity in m/s 3 marks - a 13. on a construction site, a crane is moving a ( 2.0 \times 10^{3} mathrm{~kg} ) load to the top of a building. the load starts from rest and accelerates with a uniform acceleration for the next 15 seconds. if the tension force in the cable is ( 2.0 \times 10^{4} mathrm{~n} ) up, (a) draw a free - body diagram of the load; (b) find the acceleration of the load. 1 mark - c, 3 marks - a

Explanation:

Question 12

Step1: Convert velocity unit

Given \(v = 33\ km/h\). To convert it to \(m/s\), use the conversion factor \(1\ km = 1000\ m\) and \(1\ h=3600\ s\). So \(v=\frac{33\times1000}{3600}\ m/s=\frac{330}{36}\ m/s=\frac{55}{6}\ m/s\approx9.17\ m/s\)

Step2: Use the kinematic equation \(v^{2}=u^{2}+2ad\)

Since \(u = 0\) (starts from rest), \(a=\frac{v^{2}}{2d}\). Substitute \(v=\frac{55}{6}\ m/s\) and \(d = 15\ m\). Then \(a=\frac{(\frac{55}{6})^{2}}{2\times15}=\frac{\frac{3025}{36}}{30}=\frac{3025}{1080}\ m/s^{2}\approx 2.80\ m/s^{2}\)

Step3: Use Newton's second law \(F_{net}=ma\)

Given \(m = 75\ kg\) and \(a=\frac{3025}{1080}\ m/s^{2}\). So \(F_{net}=75\times\frac{3025}{1080}=\frac{226875}{1080}\ N=\frac{15125}{72}\ N\approx210\ N\)

A free - body diagram of the load has two forces: the tension force \(T = 2.0\times10^{4}\ N\) acting upwards and the weight force \(F_{g}=mg\) acting downwards. Here \(m = 2.0\times 10^{3}\ kg\) and \(g = 9.8\ m/s^{2}\), so \(F_{g}=2.0\times10^{3}\times9.8=1.96\times10^{4}\ N\)

Step1: Use Newton's second law \(F_{net}=ma\)

The net force \(F_{net}=T - F_{g}\). Given \(T = 2.0\times10^{4}\ N\) and \(F_{g}=mg=2.0\times10^{3}\times9.8 = 1.96\times10^{4}\ N\). So \(F_{net}=2.0\times10^{4}-1.96\times10^{4}=400\ N\)

Step2: Solve for acceleration \(a\)

From \(F_{net}=ma\), we can solve for \(a\). Given \(m = 2.0\times10^{3}\ kg\) and \(F_{net}=400\ N\). Then \(a=\frac{F_{net}}{m}=\frac{400}{2.0\times10^{3}}=0.2\ m/s^{2}\)

Answer:

The net force on the athlete is approximately \(210\ N\)

Question 13 (a)