A graph of kinetic energy, in joules ( y – axis) against the square of the speed in (m / s) 2 (x – axis) is plotted for an object of mass 2.5 kg travelling along the surface of the Earth. The result is a straight line. What is the numerical value of the gradient of this line?

Answer:

27) C

We know that KE = 0.5mv²

We first need to find out whether the graph will go through zero or not.
Using the eqaution above, we can see that if v² = 0, then K.E. will also be equal to 0.
So the graph does go through 0,0.

y = mx + c
Since c (y-intercept) = 0, we get the equation,

y = mx
K.E. = 0.5 mv²

We know the K.E. is on the y-axis and the square of speed is on the x-axis.
Therefore, m (gradient) in the equation (y = mx) would be 0.5 m (0.5 mass).
We know that the mass = 2.5 kg
So gradient = 0.5 x 2.5 = 1.25