Given,
The mass of the cart, M=1.3 kg
The mass of the hanging weight, m=1.2 kg
The time instant, t=3 s
The coefficient of friction, μ=0.5
The angle of inclination of the ramp, θ=12°
The force diagram:
Where f is the frictional force acting on the cart, g is the acceleration due to gravity, and a is the acceleration of the system.
The net force acting on the cart is,
[tex]Ma=T-Mg\sin \theta-f[/tex]
On rearranging the above equation,
[tex]T=Mg\sin \theta+Ma+f\text{ }\rightarrow\text{ (i)}[/tex]
The frictional force on the cart is,
[tex]\begin{gathered} f=N\mu \\ =Mg\mu \end{gathered}[/tex]
Where N is the normal force.
On substituting the known values in the above equation,
[tex]\begin{gathered} f=1.3\times9.8\times0.5 \\ =6.37\text{ N} \end{gathered}[/tex]
The net force acting the hanging weight is given by,
[tex]ma=mg-T\text{ }\rightarrow\text{ (}ii)[/tex]
On substituting equation (i) in equation (ii),
[tex]\begin{gathered} ma=mg-Mg\sin \theta-Ma-f \\ \Rightarrow(M+m)a=mg-Mg\sin \theta-f \\ a=\frac{mg-Mg\sin \theta-f}{(M+m)} \end{gathered}[/tex]
On substituting the known values in the above equation,
[tex]\begin{gathered} a=\frac{1.2\times9.8-1.3\times9.8\times\sin 12^{\circ}-6.37}{(1.3+1.2)} \\ =1.1m/s^2 \end{gathered}[/tex]
Let us assume that the system started from the rest, thus the initial velocity of the cart is 0 m/s
From the equation of the motion,
[tex]s=ut+\frac{1}{2}at^2[/tex]
Where s is the distance traveled by the cart.
On substituting the known values in the above equation,
[tex]\begin{gathered} s=0+\frac{1}{2}\times1.1\times3^2 \\ =4.95\text{ m} \end{gathered}[/tex]
Thus the expected position of the cart after 3 seconds is 4.95 m from the bottom of the ramp.
The measured value of the position of the cart is d=5.454 m
The percent difference in the expected and the measured values is,
[tex]\begin{gathered} D=\frac{d-s}{d}\times100 \\ =\frac{5.454-4.95}{5.454}\times100 \\ =9.24\text{ percent} \end{gathered}[/tex]
Thus the percentage difference between the measured and the expected value is 9.24%
