The acceleration (in m/s 2 ) versus force (in Newtons) graph of an object has a slope of 0.5. What is the mass of the object in kilograms ?
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Firstly we know that,
y = 2 1 x ⇒ x= acceleration ( m / s 2 ), y = force ( N )
1 N = 1 k g m / s 2
and 2 N = 2 k g m / s 2
Next we must find the point like this :
If x = 2 , then y = 1
If x = 4 , then y = 2
back to the question , we must find the mass of object in kg
substitute Force = 2 N and acceleration = 1 m / s 2 , so we can rewrite
2 N = x k g m / s 2 ⇒ let x k g is mass of object
Back to The top, cause 2 N = 2 k g m / s 2 ,
so the required answer is 2 kg
F=ma AND m=f/a the slope is y/x since the force on the x-axis and the acc on the y-axis therefore the slope=y/x =a/f=1/2 so m=f/a=2 Kg
we know that slope = del y / del x so it means that acceleration / force = 1/2 thus 1/mass = 1/2 Mass = 2kg
"Mass = 2kg" is a different line
We can solve the problem fast by deriving the equation: F=ma. F/a=ma/a ----> F/a=m 2/1=m 2kg=m
Using rise/run, slope= 1/2, the numerator is acceleration and the denominator is the force. 1/2=m, hence, m=2.
We know the general straight equation for a straight line having y -intercept c and slope m as y = m x + c where x , y are the coordinate axes. Let us take acc. as y -axis and force as the x -axis. Now, y -intercept c is the value of y when x = 0 in the graph. Since, here force(F) and acc.(a) are taken as x , y axes respectively, so we have y -intercept c = 0 as without any force, i.e, x = 0 , we should have zero acc. i.e., y = 0 and thus c = 0 .
Also, we must note that since, 'F' is in Newtons and 'a' is in m / s 2 , so mass(m) should also be in S.I. i,e, have unit k g .
Now, we get the two equations of force -- (i) One using the equation y = m x + c where y = a , x = F and m = 2 1 , c = 0 and (ii) The general equation of force derived from Newton's 2nd law, F = m a . So, the two equations are --
a = 2 1 × F + 0 ⟹ F = 2 a .......(i)
F = m a ....(ii)
From equations (i) and (ii), we get ---
m a = 2 a ⟹ m = 2 kg
4N=2m/s^2 * M .... so M is 2kg
since slope= y/x; 1/2=acceleration/force; force=m a; 1/2=a/m a; m=2 kg
The slope = y/x, which is from the graph is a/F, where a: is the acceleration, and F: is the force and that is given to be 1/2 = a/F, and knowing that : F= m*a, then a/F = 1/m so 1/m = 1/2, then mass = 2 Kg.
Slope of graph gives a/F , which is equal to 1/2 . Therefore a/F = 1/2. And F/a = m . We know that a/F = 1/2 from graph, reciprocating it we get m = F/a = 2/1 = 2 kg
y2-y1/x2-x1=1/2 y1=0 x1=0 from origin so y2/x1=1/2 2*y2=x1 x1/y1=2 mass=2kg as x axis represent force and y axis acceleration
F=m a By gragh we get a/F = 1/2 Hence F = 2 a
By comparing with F = m*a. We get m = 2 kg
We understand that Force is represented as F = m a . To find m , or the mass of the object, we need to manipulate the function to isolate it.
F = m a m F = a m F ⋅ F 1 = a ⋅ F 1 m 1 = F a
Once m is isolated, we can now find what the mass of the object is. Given what we know, that the slope of a line is equal to the change in y divided by the change in x or Δ x Δ y and that the acceleration lies on the y -axis and the force on the x - axis, we can safely assume that the mass is equal to the force in this example.
m 1 = F a ⟹ m 1 = 2 1 ⟹ m = 2 .
Thus, the mass of the object is 2 kg.
F = ma
In terms of force, mass is directly proportional to acceleration. If the acceleration is 1/2 m/s^2, the mass must be 2 kg in order to preserve this direct proportionality.
1 N = (x kg)(1/2 m/s^2) x = 2 kg
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We know F = m a Also, the problem implies that F a = 2 1 ⟹ F = 2 a Finally, F = m a = 2 a