Question

as a 2 kg ball rolls down a ramp, the net force on it is 10 N. what is the acceleration?

Answer 1

The ramp height is h, and the force parallel to the ramp is half the gravitational force, 20N=(2kg)(10N/kg), so the interior angle the ramp makes with the ground is theta=30o so that sin(theta)=0.5. The ramp is thus s=2h in length. The work done will be mgh, the change in potential energy, =20N times h, which equals the work done by the component of the gravitational force parallel to the ramp, 10N times s=2h.

The energy will be divided into translational and rotational energy,

mgh = (m/2)v^2 + (I/2)(omega)^2

where omega is the angular velocity, = v/R for a sphere.
and I is the moment of inertia for a sphere, (2/5)mR^2.

Substituting and simplifying,

we have 2gh = v^2(1+2/5)

or v^2 = 2gh/(1+2/5).

For constant acceleration from zero, v^2=2as
where s is the ramp length and a the acceleration along the ramp.

Simplifying again, we have that the translational acceleration is
a=g/2[1+2/5]

where the 2 comes from the slope (20N goes to 10N)
and the [1+2/5] factor corrects for the uptake of energy by rotation.

Answer 2

Question: Since we have "the net force", does not imply that all force on the ball are calculated? like gravitational force and friction one? Why don't we apply Newton second law F = ma to find out a?

Answer 3

Yes, "net force" has some ambiguity. Presumably there is some friction present, to produce rolling rather than sliding. F=ma and F=10N is so simple that I doubt this was the intent of the question. Since it specifies rolling, we need to include rolling, which F=ma would not fully.

Answer 4

so what formula we used to determined the answer?

Answer 5

a=g/2(1+2/5)