Question

The mass of the cart is 1.3 kgThe mass of the hanging weight is 1.2 kgFor 3s the distance is. 5.454mI

Answer 1

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,

`Ma=T-Mg\sin \theta-f`

On rearranging the above equation,

`T=Mg\sin \theta+Ma+f\text{ }\rightarrow\text{ (i)}`

The frictional force on the cart is,

`\begin{gathered} f=N\mu \\ =Mg\mu \end{gathered}`

Where N is the normal force.

On substituting the known values in the above equation,

`\begin{gathered} f=1.3*9.8*0.5 \\ =6.37\text{ N} \end{gathered}`

The net force acting the hanging weight is given by,

`ma=mg-T\text{ }\rightarrow\text{ (}ii)`

On substituting equation (i) in equation (ii),

`\begin{gathered} ma=mg-Mg\sin \theta-Ma-f \\ \Rightarrow(M+m)a=mg-Mg\sin \theta-f \\ a=(mg-Mg\sin \theta-f)/((M+m)) \end{gathered}`

On substituting the known values in the above equation,

`\begin{gathered} a=(1.2*9.8-1.3*9.8*\sin 12^(\circ)-6.37)/((1.3+1.2)) \\ =1.1m/s^2 \end{gathered}`

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,

`s=ut+(1)/(2)at^2`

Where s is the distance traveled by the cart.

On substituting the known values in the above equation,

`\begin{gathered} s=0+(1)/(2)*1.1*3^2 \\ =4.95\text{ m} \end{gathered}`

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,

`\begin{gathered} D=(d-s)/(d)*100 \\ =(5.454-4.95)/(5.454)*100 \\ =9.24\text{ percent} \end{gathered}`

Thus the percentage difference between the measured and the expected value is 9.24%