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A cart moving at 10 m/s is brought to a stop by the force plotted in the force-time graph shown here. Find the impulse and the approximate mass of the cart. Show all your work.

Having a lot of trouble with this question. If anyone answers me could you tell me how to solve the question and how you manage to find approximate mass, when you aren't given the velocity? Thank you

A cart moving at 10 ms is brought to a stop by the force plotted in the forcetime graph shown here Find the impulse and the approximate mass of the cart Show al class=

Respuesta :

Answer:

Impulse = 88 kg m/s

Mass = 8.8 kg

Explanation:

We are given a graph of Force vs. Time. Looking at the graph we can see that the Force acts approximately between the time interval from 1sec to 4sec.

Newton's Second Law relates an object's acceleration as a function of both the object's mass and the applied net force on the object. It is expressed as:

[tex]F=ma[/tex]      Eqn. (1)

where

[tex]F[/tex] : is the Net Force in Newtons ( [tex]N[/tex] )

[tex]m[/tex] : is the mass ( [tex]kg[/tex] )

[tex]a[/tex]  : is the acceleration ( [tex]m/s^2[/tex] )

We also know that the acceleration is denoted by the velocity ( [tex]v[/tex] ) of an object as a function of time ( [tex]t[/tex] ) with

[tex]a=\frac{v}{t}[/tex]         Eqn. (2)

Now substituting Eqn. (2) into Eqn. (1) we have

[tex]F=m\frac{v}{t}\\ \\Ft=mv[/tex]     Eqn. (3)

However since in Eqn. (3) the time-variable is present, as a result the left hand side (i.e. [tex]Ft[/tex] is in fact the Impulse  [tex]J[/tex] of the cart ), whilst the right hand side denotes the change in momentum of the cart, which by definition gives as the impulse. Also from the graph we can say that the Net Force is approximately ≈ [tex]22N[/tex] and [tex]t=4 sec[/tex] (thus just before the cut-off time of the force acting).

Thus to find the Impulse we have:

[tex]J=Ft\\J=(22N)(4sec)\\J=88 kg m/s[/tex]

So the impulse of the cart is [tex]J=88kg m/s[/tex]

Then, we know that the cart is moving at [tex]v=10 m/s[/tex]. Plugging in the values in Eqn. (3) we have:

[tex](22N)(4sec)=(10m/s)m\\\\88=10m\\\\m=\frac{88}{10}\\ \\m=8.8kg[/tex]

So the mass of the cart is [tex]m=8.8kg[/tex].

The required value of impulse is 88 N-s and the mass of the cart is 8.8 kg.

Newton's Second Law relates an object's acceleration as a function of both the object's mass and the applied net force on the object.

Given data:

The speed of cart is, v = 10 m/s.

We are given the Force-time graph for which the area of the force-time graph will provide the value of impulse. So from the graph, the maximum force is,

F = 22 N

And time interval is,

t = 4.5 s - 0.5 s

t = 4 s

So, the impulse is,

I = Area of graph

I = F × t

I = 22 × 4

I = 88 N-s

Now the standard expression for the impulse as per the impulse-momentum theorem is,

I = ΔP (Change in momentum)

I = m ×( v - u )

Here, u is the final velocity, and since the cart stops finally, then u = 0 m/s. And m is the mass of the cart.

Solving as,

88 = m ×( v - u )

88 = m ×( 10 - 0 )

m = 8.8 kg

Thus, we can conclude that the required value of impulse is 88 N-s and the mass of car is of 8.8 kg.

Learn more about the impulse here:

https://brainly.com/question/904448