Answer:
The answer is below
Explanation:
Given that:
mass (m) = 86 kg, distance (L) = 2.75 m, θ = 31°, force (F) = 595 N, initial velocity ([tex]v_i[/tex]) = 2.4 m/s, g = acceleration due to gravity = 9.8 m/s²
The net work can be gotten from the equation:
[tex]W_{net}=Fcos(\theta)L-mgsin(\theta)L\\\\W_{net}=595*cos(31)*2.75-[86*9.81*sin(31)*2.75)\\\\W_{net}=1402.54-1194.92\\\\W_{net}=207.62[/tex]
From the work-energy theorem equation, we can get her speed at the top of the ramp ([tex]v_f[/tex])
Hence:
[tex]W_{net}=change\ in\ kinetic\ energy\\\\W_{net}=\frac{1}{2}m(v_f^2-v_i^2 )\\\\2W_{net}=m(v_f^2-v_i^2 )\\\\v_i=\sqrt{ v_i^2+\frac{2W_{net}}{m}} \\\\v_f=\sqrt{ 2.4^2+\frac{2*207.62}{86}}}\\\\v_f=3.25\ m/s[/tex]
Using the concept of work - energy theorem, Dr. Moreno's speed at the top of the ramp is obtained as 3.25 m/s.
This problem can be analysed using a free-body diagram.
We know that work done by a force is given by;
[tex]W= F.d[/tex]
So, the net work done in the horizontal direction is given by;
[tex]W_{net}= Fcos\theta.d-mgcos \theta.d[/tex]
[tex]W_{net}= (595\,N\times cos\,31^\circ \times 2.75\,m)-(86 \times 9.8\, m/s^2cos\,31^\circ \times 2.75\,m)[/tex]
[tex]\implies W_{net}=1402.539\,N - 1193.703\,N = 208.836\,N[/tex]
According to the Work-Energy theorem, the net work done is the change in Kinetic energy.
[tex]W = (KE)_f - (KE)_i[/tex]
[tex]W = \frac{1}{2} m (v_f)^2 -\frac{1}{2} m (v_i)^2[/tex]
[tex]v_f = \sqrt{\frac{2W}{m} + (v_i)^2}[/tex]
Substituting the known values, we get;
[tex]v_f = \sqrt{\frac{2 \times 208.836}{86\,kg}+ (2.4\,m/s)^2}=3.25\,m/s[/tex]
Learn more about work-energy theorem here:
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